first principle of differentiation pdf

A tangent touches the curve at one point, and the gradient varies according to the touching coordinate. 2. Answer sheets of meritorious students of class 12th 2012 M.P Board All Subjects. For different pairs of points we will get different lines, with very different gradients. We begin by looking at the straight line. << /S /GoTo /D [2 0 R /Fit] >> U)dFQPQK$T8D*IRu"G?/t4|%}_|IOG$NF\.aS76o:j{ I am trying to differentiate the functions x n, e ax and ln (ax) from first principles. 2.00 4.00 6.00 8.00 100 200 300 (metres) Distance time (seconds) Mathematics Learning Centre, University of Sydney 1 1 Introduction In day to day life we are often interested in the extent to which a change in one quantity Regrettably mathematical and statistical content in PDF les is unlikely to be l]*-.[p-x$CII L?& gM=:?b.pB>= ! The points A and B lie on the curve and have x-coordinates 5 and 5-+11 respectively, where h > O. Sign In, Create Your Free Account to Continue Reading, Copyright 2014-2021 Testbook Edu Solutions Pvt. Follow the following steps to find the derivative of any function. 2 3 * 2 2 = 2*2*2*2*2 = 2 5 x m / x n = x m-n eg. This article explains the first principle of differentiation which states that the derivative of a function $f(x)$ with respect to $x$ and it is given by ${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$. 2-3 = 2 1 / 2 4 2/16 = 1/8 x m x 1/n = nx eg. Plugging \sqrt{x} into the definition of the derivative, we multiply the numerator and denominator by the conjugate of the numerator, \sqrt{x+h}+\sqrt{x}. > Differentiating powers of x. Differentiate ${\sin ^2}x$ with respect to $x$ from first principle.Ans: Given: $f(x) = {\sin ^2}x$$ \Rightarrow f(x + h) = {\sin ^2}(x + h)$${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{{{\sin }^2}(x + h) {{\sin }^2}x}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sin (x + h + x)\sin (x + h x)}}{h}\,\,\,\,\,\,\left[ {{{\sin }^2}A {{\sin }^2}B = \sin (A + B)\sin (A B)} \right]$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sinh }}{h} \times \mathop {\lim }\limits_{h \to 0} [\sin (2x + h)]$$ = 1(\sin 2x)$$\therefore \frac{d}{{dx}}(x) = \sin 2x$, Q.4. Ltd.: All rights reserved, Definition of First Principles of Derivative. Differentiation from First Principles The formal technique for finding the gradient of a tangent is known as Differentiation from First Principles. Sharma vs S.K. The "first principle" is the Fundamental Theorem of Calculus, which proves the definite integral / Riemann sum (which Mandelbroth gave) is equal to where . [4] 2. If you have any doubts, then do let us know about it in the comment section below. You choose the most cost-effective option according to your priority objectives (e.g., reducing GHG emissions, alleviating energy poverty, reducing local air pollution, improving indoor air quality, ensuring security . The premise of this is that the derivative of a function is the the gradient of the tangent of the function at a singular point. Q.2. Efficiency First is an elementary principle: you can influence both demand and supply in balancing the two in any single moment. Differentiate $x{e^x}$ from first principles.Ans: Given: $f(x) = x{e^x}$$ \Rightarrow f(x + h) = (x + h){e^{(x + h)}}$From the definition of first principles, we have,${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h){e^{x + h}} x{e^x}}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left( {x{e^{x + h}} x{e^x}} \right) + h{e^{x + h}}}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \left\{ {x{e^x}\left( {\frac{{{e^h} 1}}{h}} \right) + {e^{x + h}}} \right\}$$ = x{e^x}\mathop {\lim }\limits_{h \to 0} \left( {\frac{{{e^h} 1}}{h}} \right) + \mathop {\lim }\limits_{h \to 0} {e^{x + h}}$$ = x{e^x} + {e^x}\quad \left[ {\mathop {\lim }\limits_{h \to 0} \left( {\frac{{{e^h} 1}}{h}} \right) = 1} \right]$$\therefore \frac{d}{{dx}}\left( {x{e^x}} \right) = x{e^x} + {e^x}$$ \Rightarrow \frac{d}{{dx}}\left( {x{e^x}} \right) = {e^x}(x + 1)$, Q.5. Get Daily GK & Current Affairs Capsule & PDFs, Sign Up for Free The first principle of differentiation is to compute the derivative of the function using the limits. We hope that this detailed article on the First Principle of Differentiation was helpful. endstream endobj 203 0 obj <>/Metadata 8 0 R/Outlines 12 0 R/PageLayout/OneColumn/Pages 200 0 R/StructTreeRoot 21 0 R/Type/Catalog>> endobj 204 0 obj <>/ExtGState<>/Font<>/XObject<>>>/Rotate 0/StructParents 0/Type/Page>> endobj 205 0 obj <>stream First Principle of Differentiation: Derivative as a Rate Measurer, Geometrical Interpretation of Derivative at a Point A derivative is the first of the two main tools of calculus (the second being the integral). Let y = f(x) be a function of x. 2 3 / 2 2 = 2*2*2 = 2 1 = 2 2*2 (x m) n =x mn eg. Differentiation from first Principle: SOME EXAMPLES f '(x) = lim h0 f ( x + h ) f ( x) h is What is the first principle of differentiation?Ans: The first principle rule of differentiation helps us evaluate the derivative of a function using limits. Resources - Worksheet questions the same as PowerPoint, including the. Step 1: Click on the Download PDF button. hbbd``b`z$X3^ `I4 fi1D %A,F R$h?Il@,&FHFL 5[ How To Method of Differentiation Notes PDF? In this unit we look at how to dierentiate very simple functions from rst principles. Differentiation, in calculus, can be applied to measure the function per unit change in the independent variable. As the distance between x and x+h gets smaller, the secant line that weve shown will approach the tangent line representing the functions derivative. You need the best 9th CBSE study materials to score well in the exam. Solution: Using first principles,1 1 You need to know the identity (a +b) 2 . Further, derivative of $f$ at $x = a$ is denoted by,${\left. Did you know that more than 21 lakh students appear every year for the CBSE Class 10 exam? \(m_{tangent}=\lim _{h{\rightarrow}0}{y\over{x}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}$. Taking limit as $Q \to P$i.e. After reading this text, and/or viewing the video tutorial on this topic, you should be able to: dierentiate the function sinx from rst principles %PDF-1.5 Step 2: On that topic page click on save button. Now, what CBSE Class 9 exam is the foundation stone for your higher classes. w0:i$1*[onu{U 05^Vag2P h9=^os@# NfZe7B The play stealer works off what's already been done. The derivative of \sqrt{x} can also be found using first principles. Does differentiation give gradient? (4) A curve has equation y = 2x2. In finding the limit in each problem, you need to first Taylor expand to remove x from the denominator. Sometimes ${f^\prime }(x)$ is denoted by $\frac{d}{{dx}}(f(x))$ or if $y = f(x)$, it is denoted by $\frac{{dy}}{{dx}}$. First Derivative Calculator - Symbolab Solutions Graphing Practice New Geometry Calculators Notebook Sign In Upgrade en Pre Algebra Algebra Pre Calculus Calculus Functions Linear Algebra Trigonometry Statistics Physics Chemistry Finance Economics Conversions First Derivative Calculator Differentiate functions step-by-step Derivatives It is crucial to pay full attention while preparing for CBSE Class 8 exam, and a strong base helps create a strong foundation. Differentiate $\sqrt {2x + 3} $ with respect to $x$ from the first principle.Ans: Given: $f(x) = \sqrt {2x + 3} $$ \Rightarrow f(x + h) = \sqrt {2(x + h) + 3} $${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {2(x + h) + 3} \sqrt {2x + 3} }}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sqrt {2(x + h) + 3} \sqrt {2x + 3} } \right]\left[ {\sqrt {2(x + h) + 3} + \sqrt {2x + 3} } \right]}}{{h\left[ {\sqrt {2(x + h) + 3} + \sqrt {2x + 3} } \right]}}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{(2x + 2h + 3 2x 3)}}{h} \times \frac{1}{{\left[ {\sqrt {2x + 2h + 3} + \sqrt {2x + 3} } \right]}}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{2h}}{h} \times \mathop {\lim }\limits_{h \to 0} \frac{1}{{\left[ {\sqrt {2x + 2h + 3} + \sqrt {2x + 3} } \right]}}$$ = 2 \times \frac{1}{{\sqrt {2x + 3} + \sqrt {2x + 3} }}$$ = \frac{2}{{2(\sqrt {2x + 3} )}}$$\therefore \,\frac{d}{{dx}}(\sqrt {2x + 3} ) = \frac{1}{{\sqrt {2x + 3} }}$, Q.2. > Differentiating logs and exponentials. We begin by looking at the straight line. This is known as the average rate of change of $y$ with respect to $x$.As $\Delta x \to 0$, we observe that $\Delta y \to 0$. First Principles of Derivatives are useful for finding Derivatives of Algebraic Functions, Derivatives of Trigonometric Functions, Derivatives of Logarithmic Functions. But the very process of Taylor expansion uses differentiation to find its coefficients. Thus, the derivative of a function $f(x)$ at a point $x= c$ is the slope of the tangent to the curve $y = f(x)$ at the point $(c,f(c))$. > Differentiation from first principles. Out of these, nearly 19 lakh students manage to pass the exam, but only 5 lakh students score above 90%. 2ax. Download Now! > Using a table of derivatives. View a short video on differentiation from first principles. Contents: PowerPoint - What is differentiation?, using DESMOS. This method is called differentiation from first principles or using the definition. Now lets see how to find out the derivatives of the trigonometric function. We will derive the derivative of sin x using the first principle of differentiation, that is, using the definition of limits. Optional Investigation Rules for differentiation Differentiate the following from first principles: f (x) = x f ( x) = x f (x) = 4x f ( x) = 4 x f (x) = x2 f ( x) = x 2 There are various methods of differentiation. {\frac{{dy}}{{dx}}} \right|_{x = 2}} = 2(2) 2 = 2\)Hence, the slope of the given curve at the given point is $2$.Thus, the slope of a curve at a point is found using the first derivative. /Filter /FlateDecode hb```+@(1P,rl @ @1C .pvpk`z02CPcdnV\ D@p;X@U 8 0 obj Thus, we observe that due to change $\Delta x$ in $x$, there is a change $\Delta y$ in $y$. Prove, from first principles, that the derivative of kx3 is 3kx2. Let $\Delta x$ be a small change (positive or negative) in $x$ and let $\Delta y$ be the corresponding change in $y = f(x)$. But it's essential that we show them where the rules come from, so let's look at that. This research work will give a vivid look at differentiation and its application. Differentiation from First Principles Save Print Edit Differentiation from First Principles Calculus Absolute Maxima and Minima Accumulation Function Accumulation Problems Algebraic Functions Alternating Series Antiderivatives Application of Derivatives Approximating Areas Arc Length of a Curve Arithmetic Series Average Value of a Function Each is the reverse process of the other. The indefinite integral of is defined as the antiderivative of (plus a generic constant), by analogy with the Fundamental Theorem. {\frac{{dy}}{{dx}}} \right|_{x = c}}\), which is also the formula to find the gradient of the curve the point $(c,f(c))$.For Example: The slope of the curve $y = {x^2} 2x 3$ at the point $P(2, 3)$ is evaluated as follows:$\frac{{dy}}{{dx}} = 2x 2$And, at $x= 2$, we have${\left. DHNR@ R$= hMhNM Dierentiate y=(2x+1)3(x8)7 with respect to x. (3 marks) (4 marks) (4 marks) f(x) = ax2, where a is a constant. x 1/2 * x 1/2 = x 1 Therefore x 1 = x 1/2 x m/n = nx m. Sorry, preview is currently unavailable. For f(a) to exist it is necessary and sufficient that these conditions are met: Furthermore, if these conditions are met, then the derivative f (a) equals the common value of \(f_{-}(a)\text{ and }f_{+}(a)$ i.e. $f(a)=f_{-}(a)=f_{+}(a)$, $\begin{matrix} f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{f(-7+h)f(-7)\over{h}}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{|(-7+h)+7|-0\over{h}}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{|h|\over{h}}\\ \text{as h < 0 in this case}\\ f_{-}(-7)=\lim _{h{\rightarrow}{0^-}}{-h\over{h}}\\ f_{-}(-7)=-1\\ \text{On the other hand}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{f(-7+h)f(-7)\over{h}}\\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{|(-7+h)+7|-0\over{h}}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{|h|\over{h}}\\ \text{as h > 0 in this case}\\ f_{+}(-7)=\lim _{h{\rightarrow}{0^+}}{h\over{h}}\\ f_{+}(-7)=1\\ \therefore{f_{-}(a)\neq{f_{+}(a)}} \end{matrix}$, Therefore, f(x) it is not differentiable at x = 7. o}-ixuYF^+@-l ,:2cG^7OeE1?lEHrS,SvDW B6M7,-;g ,(e_ZmeP. Answer (1 of 2): I'm going top assume you mean differentiation from first principles. Derivative by the first principle is also known as the delta method. endstream endobj startxref ZL$a_A-. Differentiation by first principles refers to find a general expression for the slope or gradient of a curve using algebraic techniques. It will state the fundamental of calculus, it shall also deal with limit and continuity. 1 0 obj X)YJ*D]R**j,)'N DYrf:lx|6 I have successful in all three, but here's my problem. Then,Slope of chord, $PQ = \tan \angle QPN$$ = \frac{{QN}}{{PN}}$$\therefore \,PQ = \frac{{f(c + h) f(c)}}{h}$. You will be taken to download page. The basic principle of integration is to reverse differentiation. Differentiation from first principles of some simple curves For any curve it is clear that if we choose two points and join them, this produces a straight line. Answer. Academia.edu no longer supports Internet Explorer. How to Find a Derivative using the First Principle? Our team will get try to solve your queries at the earliest. 6: The Quotient . We know that, $f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}$. It means that the slope of the tangent line is equal to the limit of the difference quotient as h approaches zero. An integral is sometimes referred to as antiderivative. Q.4. Where can you You must be surprised to know that around 2M+ students appear for the CBSE Class 10 exams every year! For this work to be effectively done, there is need for the available of time, important related text book and financial aspect cannot be left out. Show, from first principles, that the derivative of 3x2 is 6x so STEP 2: Expand f (x+h) in the numerator STEP 3: Simplify the numerator, factorise and cancel h with the denominator STEP 4: Evaluate the remaining expression as h tends to zero 5: The Product Rule Pt. Example: The derivative of a displacement function is velocity. Worked example 7: Differentiation from first principles Calculate the derivative of g ( x) = 2 x 3 from first principles. So, change in the value of $f$ is given by,$f(x + \Delta x) f(x)$ or $\Delta y = f(x + \Delta x) f(x)\quad \ldots \ldots (i)$. This module provides some examples on differentiation from first principles. [5] (b) Given that and when x = 4, find the value of the constant a. It is also known as the delta method. y = f(x), then the proportional x = y. dx dy 1 = dx d (ln y ) Take logs and differentiate to find proportional changes in variables Differentiate $x{\tan ^{ 1}}x$ from first principleAns: $f(x) = x{\tan ^{ 1}}x$, then $f(x + h) = (x + h){\tan ^{ 1}}(x + h)$Using the first principle of differentiation,${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{(x + h){{\tan }^{ 1}}(x + h) x{{\tan }^{ 1}}x}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \left\{ {x\left\{ {\frac{{{{\tan }^{ 1}}(x + h) {{\tan }^{ 1}}x}}{h}} \right\} + \frac{{h{{\tan }^{ 1}}(x + h)}}{h}} \right\}$$ = \mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{x{{\tan }^{ 1}}\left( {\frac{{x + h x}}{{1 + x(x + h)}}} \right)}}{h}} \right\} + \mathop {\lim }\limits_{h \to 0} {\tan ^{ 1}}(x + h)$$\left[ {\because {{\tan }^{ 1}}x {{\tan }^{ 1}}y = {{\tan }^{ 1}}\left( {\frac{{x y}}{{1 + xy}}} \right);xy > 1} \right]$$ = x\mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{{{\tan }^{ 1}}\left( {\frac{h}{{1 + x(x + h)}}} \right)}}{{\frac{h}{{1 + x(x + h)}}}} \times \frac{1}{{1 + x(x + h)}}} \right\} + {\tan ^{ 1}}x$$ = \frac{x}{{1 + {x^2}}} + {\tan ^{ 1}}x$Hence, $\frac{d}{{dx}}\left( {x{{\tan }^{ 1}}x} \right) = \frac{x}{{1 + {x^2}}} + {\tan ^{ 1}}x$. Consider the curve $y = f(x)$.Let $P(c,f(c))$ be a point on the curve $y = f(x)$ and let $Q(c + h,f(c + h))$ be a neighbouring point on the same curve. sF1MOgSwEyw1zVt'B0zyn_'sim|U.^LV\#.=F?uS;0iO? 4: The Chain Rule Pt. +6r6aM^clsq@y)X)1hG!*8"Qebo4Aa`'V&aU!B(AAFbDFL |%/e&RC%0Ka`UOLZob"MlM) You'll notice that all but one of the terms contains an . We illustrate this in Figure 2. Here, $x$ is the independent variable, and $y$ is the dependent variable on $x$. A generalization of the concept of a derivative, in which the ordinary limit is replaced by a one-sided limit. neB;!U~^Umt89[d5pNGt"9Hvk)&hyJwCY1UGmTA[M4U1MR[{2vt1Be' Pw6U\l( S?IT :+P Differentiation of the sine and cosine functions from rst principles In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. This method is called differentiation from first principles or using the definition. "J;m*;H@|V, 0;sMrZqVP-Eaz0!. Pt. Where k is a constant. Differentiation of Trigonometric Functions using First Principles of Derivatives, Derivative of sinx by the first principle, Derivative of cosx by the first principle, Derivative of tanx by the first principle, d-Block Elements: Periodic, Physical Properties and Chemical Properties, Parts of Circle : Learn Definition with Properties, Formula and Diagrams, Applications of VSEPR Theory, Examples with Answers and Explanations, Matrix Addition: Meaning, Properties, How to add with Solved Examples, Polar Form of Complex Numbers with Equations in Different Quadrants using Solved Examples. This is the same thing as the slope of the tangent line to the graph of the function at that point. The tangents of the function f (x)=x can be explored using the slider below. In this unit we look at how to dierentiate very simple functions from rst principles. Enter the email address you signed up with and we'll email you a reset link. 244 0 obj <>stream We denote derivatives as ${dy\over{dx}}\($, which represents its very definition. Going off of the basis that gradient is the change in y over an interval in x, t. The derivative is a measure of the instantaneous rate of change, which is equal to: $f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}$. Differentiating from First Principles SOCUTLONS Differentiating from First Principles - Edexcel Past Exam Questions (a) Given that y = 2x2 5x+3, find A from first principles. ) In general, you need to know a bit of algebra to do limits effectively. Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. - Examples of Differentiation from first principles, easy, medium and difficult. $h \to 0$, we get,$\mathop {\lim }\limits_{Q \to P} $(Slope of chord$PQ$) $ = \mathop {\lim }\limits_{h \to 0} \frac{{f(c + h) f(c)}}{h} \ldots . (ii)$As $Q \to P$, chord $PQ$ tends to the tangent to $y = f(x)$ at point $P$.Therefore, from $(ii)$, we haveSlope of the tangent at $P = \mathop {\lim }\limits_{h \to 0} \frac{{f(c + h) f(c)}}{h}$$ \Rightarrow $Slope of the tangent at $P = {f^\prime }(c)$ i.e., $\tan \theta = {f^\prime }(c)$, where $\theta$ is the inclination of the tangent to the curve $y = f(x)$ at point $(c,f(c))$ with the $x$axis. Let a function of a curve be y = f (x). We illustrate below. The derivative of a function is simply the slope of the tangent line that passes through the functions curve. Solucionario en Ingls del libro "Clculo: Trascendentes tempranas" del autor Dennis G. Zill, n = x m+n eg. The derivative of a function, represented by ${dy\over{dx}}$ or f(x), represents the limit of the secants slope as h approaches zero. Differentiating a linear function A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. $\therefore $ Rate of change in $y$ with respect to $x = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\Delta y}}{{\Delta x}}$.$ = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) f(x)}}{{\Delta x}}$ [Using $(i)$]$ = \frac{d}{{dx}}(f(x))$ [Using definition of first principleof differentiation]$ = \frac{{dy}}{{dx}}$, Thus, $ \frac{{dy}}{{dx}}$ measures the rate of change of $y = f(x)$ with respect to $x$.i.e., $\frac{{dy}}{{dx}} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{f(x + \Delta x) f(x)}}{{\Delta x}}$, Conclusion: We can say that the derivative of a function $y = f(x)$ is same as the rate of change of $f(x)$ with respect to $x$, Let $f(x)$ be a differentiable function. Goyal, Mere Sapno ka Bharat CBSE Expression Series takes on India and Dreams, CBSE Academic Calendar 2021-22: Check Details Here. Click on each book cover to see the available files to download, in English and Afrikaans. For a linear function this is a trivial exercise because the graph of the function is a straight line. Check out this article on Limits and Continuity. rst principles mc-TY-sincos-2009-1 In order to master the techniques explained here it is vital that you undertake plenty of practice exercises so that they become second nature. The derivative of a function $y = f(x)$ is same as the rate of change of $f(x)$ with respect to $x$. Sure, maybe he adds a tweak here or there, but by and large he's just copying something that someone else created. The derivative can also be represented as f(x) as either f(x) or y. Let us take a point P with coordinates (x, f (x)) on a curve. First Principles Once students start differentiating using a set of rules, this topic is fairly straightforward. > Differentiating sines and cosines. The approach is practical rather than purely mathematical and may be too simple for those who prefer pure maths. sfujFKZ(**s/B '2M(*G*iB B,' gvW$ After reading this text, and/or viewing the video tutorial on this topic, you should be able to: A derivative is the first of the two main tools of calculus (the second being the integral). EXERCISES IN MATHEMATICS, G1 Then the derivative of the function is found via the chain rule: dy dx = dy du du dx = 1 2x2 p 1+x1 Products and Quotients 7. However, we can use this method of finding the derivative from first principles to obtain rules which make finding the derivative of a function much simpler. Q.1. Example Consider the straight line y = 3x +2 shown in . $\frac{d}{{dx}}(k) = 0$, where $k$ is a constant.Power rule: $\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n 1}}$, where $n$ is any real number.Sum and Difference Rule: If $f(x) = g(x) + h(x)$ then ${f^\prime }(x) = {g^\prime }(x) + {h^\prime }(x)$. At any point on a curve, the gradient is equal to the gradient of the tangent at that point (a tangent to a curve is a line touching the curve at one point only). If the following limit exists for a function f of a real variable x: $f(x)=\lim _{x{\rightarrow}{x_o+0}}{f(x)f(x_o)\over{x-x_o}}$, then it is called the right (respectively, left) derivative of ff at the point x0x0. First Principles of Derivatives refers to using algebra to find a general expression for the slope of a curve. Differentiation from first principles Watch on Transcript Example 1 If f(x) = x2, find the derivative of f(x) from first principles. For different pairs of points we will get different lines, with very different gradients. But what if you get everything Class 8 is the foundation of any student's career. Prove, from first principles, that the derivative of 6x is 6. By using our site, you agree to our collection of information through the use of cookies. The rules of football are the first principles: they govern what you can and can't do. (2 3) 2 = (2*2*2)*(2*2*2) = 2 6 x 0 = 1, x0 eg. {\frac{{df}}{{dx}}} \right|_{x = a}}or{\mkern 1mu} {\mkern 1mu} {\left( {\frac{{df}}{{dx}}} \right)_{x = a}}\). Procedure for CBSE Compartment Exams 2022, Maths Expert Series : Part 2 Symmetry in Mathematics. This is called as First Principle in Calculus. Both $f_{-}(a)\text{ and }f_{+}(a)$ must exist. What is Differentiation in Maths. Prove, from first principles, that the derivative of 3x2 is 6x. 3: General Differentiation Pt. Application III: Differentiation of Natural Logs to find Proportional Changes The derivative of log(f(x)) f'(x)/ f(x), or the proportional change in the variable x i.e. They are a part of differential calculus. Solution: Using first principles, 1 1 You need to know the identity (a + b)2 = a2 + 2ab + b2 for this example. View Differentiation From First Principles (1).pdf from MAT CALCULUS at University of South Africa. Leading AI Powered Learning Solution Provider, Fixing Students Behaviour With Data Analytics, Leveraging Intelligence To Deliver Results, Exciting AI Platform, Personalizing Education, Disruptor Award For Maximum Business Impact, First Principle of Differentiation: Derivative as a Rate Measurer, Geometrical Interpretation of Derivative at a Point, All About First Principle of Differentiation: Derivative as a Rate Measurer, Geometrical Interpretation of Derivative at a Point. Differentiating a linear function A straight line has a constant gradient, or in other words, the rate of change of y with respect to x is a constant. In Mathematics, Differentiation can be defined as a derivative of a function with respect to an independent variable. Then, as the value of $x$ changes from $x$ to $x + \Delta x$ and the value of $f(x)$ changes from $f(x)$ to $f(x + \Delta x)$. There are different notations for derivative of a function. Now PQ is the secant to the curve. Tutorials in differentiating logs and exponentials, sines and cosines, and 3 key rules explained, providing excellent reference material for undergraduate study. Please note that the PDF may contain references to other parts of the module and/or to software or audio-visual components of the module. Suppose $f$ is a real valued function, the function defined by$\mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$, wherever the limit exists is defined to be the derivative of $f$ at $x$ and is denoted by ${f^\prime }(x)$.This definition of derivative is called the first principle of differentiation.$\therefore \,{f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$. Get some practice of the same on our free Testbook App. Q.5. What are the three rules of differentiation?Ans: The three rules of differentiation areConstant rule: The constant rule states that the derivative of a constant is zero i.e. Everything is possible as long as it's not against the rules. The tangent to x^2 slider 2. 2 2 = 1 2 2 = 2 0 2 2 2 2 x-m = 1 eg. To browse Academia.edu and the wider internet faster and more securely, please take a few seconds toupgrade your browser. Let $f(x)$ be a function of $x$ and let $y = f(x)$. Section 12.1 Instructor's Resource Manual CHAPTER 12 Derivatives for Functions of Two or More Variables, Single Variable Calculus Early Transcendentals Complete Solutions Manual, Core Mathematics C1 Rules of Indices x m * x. You can download the paper by clicking the button above. A thorough understanding of this concept will help students apply derivatives to various functions with ease. Differentiate ${e^{\sqrt {\tan x} }}$ from first principle.Ans: Let $f(x) = {e^{\sqrt {\tan x} }}$$f(x + h) = {e^{\sqrt {\tan (x + h)} }}$From the first principle${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{{e^{\sqrt {\tan \left( {x + h} \right)} }} {e^{\sqrt {^{\tan x}} }}}}{h}$$ = \mathop {\lim }\limits_{h \to 0} {e^{\sqrt {\tan x} }}\left\{ {\frac{{{e^{\sqrt {\tan (x + h)} \sqrt {\tan x} }} 1}}{h}} \right\}$$ = {e^{\sqrt {\tan x} }}\mathop {\lim }\limits_{h \to 0} \left\{ {\frac{{{e^{\sqrt {\tan (x + h)} \sqrt {\tan x} }} 1}}{{\sqrt {\tan (x + h)} \sqrt {\tan x} }} \times \frac{{\sqrt {\tan (x + h)} \sqrt {\tan x} }}{h}} \right\}$$ = {e^{\sqrt {\tan x} }} \times 1 \times \mathop {\lim }\limits_{h \to 0} \left( {\frac{{\sqrt {\tan (x + h)} \sqrt {\tan x} }}{h} \times \frac{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}} \right)\quad \left[ {\mathop {\because \lim }\limits_{x \to 0} \frac{{{e^x} 1}}{x} = 1} \right]$$ = {e^{\sqrt {\tan x} }}\mathop {\lim }\limits_{h \to 0} \frac{{\tan (x + h) \tan x}}{h} \times \frac{1}{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}$$ = {e^{\sqrt {\tan x} }}\mathop {\lim }\limits_{h \to 0} \left( {\frac{{\frac{{ \sin (x + h)}}{{ \cos (x + h)}} \frac{{ \sin x}}{{{\mathop{\rm cos}\nolimits} x}}}}{h} \times \frac{1}{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}} \right)$$ = {e^{\sqrt {\tan x} }}\mathop {\lim }\limits_{h \to 0} \left( {\frac{{ \sin (x + h) \cos x \sin x \cos (x + h)}}{{h\cos (x + h){\mathop{\rm cos}\nolimits} x}} \times \frac{1}{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}} \right)$$ = {e^{\sqrt {\tan x} }}\mathop {\lim }\limits_{h \to 0} \left( {\frac{{ \sin (x + h x)}}{{h \cos (x + h){\mathop{\rm cos}\nolimits} x}} \times \frac{1}{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}} \right)$$ = {e^{\sqrt {\tan x} }}\mathop {\lim }\limits_{h \to 0} \left( {\frac{{ \sin h}}{{h \cos (x + h){\mathop{\rm cos}\nolimits} x}} \times \frac{1}{{\sqrt {\tan (x + h)} + \sqrt {\tan x} }}} \right)$$ = {e^{\sqrt {\tan x} }}\left( {1 \times x \times \frac{1}{{2\sqrt {\tan x} }}} \right)$$ = \frac{{{e^{\sqrt {\tan x} }}}}{{2\sqrt {\tan x} }}{\sec ^2}x$Hence, $\frac{d}{{dx}}\left( {{e^{\sqrt {\tan x} }}} \right) = \frac{{{e^{\sqrt {\tan x} }}}}{{2\sqrt {\tan x} }}{\sec ^2}x$, Q.7. Prove, from first principles, that the derivative of 4x2 is 8x. [Kkb{8C_`I3PJ*@;mD:`x$QM+x:T;Bgfn It is also known as the delta method. They apply a simple procedure and get the answers right - hey presto, they're doing calculus. Q.3. What is $\frac{{dy}}{{dx}}$?Ans: $\frac{{dy}}{{dx}}$ is an operation which indicates the differentiation of $y$ with respect to $x$. [2] 3. Open Textbooks | Siyavula Open Textbooks Download our open textbooks in different formats to use them in the way that suits you. 1: First Principles 1. Study materials also help you to cover the entire syllabus efficiently. Contents [ show] We know that the gradient of the tangent to a curve with equation y = f(x) at x = a can be determine using the formula: Gradient at a point = lim h 0f(a + h) f(a) h. We can use this formula to determine an expression that describes the gradient of the graph (or the gradient of the . Further, some standard formulas of differentiation (or derivatives) of trigonometric and polynomial functions were derived using the first principle. What is the derivative of $2x$?Ans: Let $y=2x$Then, $\frac{d}{{dx}}\left( {2x} \right) = 2\frac{d}{{dx}}\left( x \right) = 2,\,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,\frac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n 1}},n \in R} \right]$Hence, the derivative of $2x$ is $2$. This is the fundamental definition of derivatives. It is theinstantaneous rate of change of a function at a point in its domain. The PDF of this extract thus shows the content exactly as it would be seen by an Open University student. The derivative of a function, represented by d y d x or f (x), represents the limit of the secant's slope as h approaches zero. Embiums Your Kryptonite weapon against super exams! endobj stream This principle is the basis of the concept of derivative in calculus. 34. << The tangent line is the result of secant lines having a distance between x and x+h that are significantly small and where h0. Prove, from first principles, that f'(x) = 2*'TD QM>K6YN3VFrs%BaF50 D~c|ULYG{$[Je& 2lI8JO sERUa6QI`qdDPo'Fds1],jsx]SuOuaO%S2>\7MELtJfMhiYRNaSmcWI)QtLLtqru from rst principles) 31 6 Solutions to exercises 35. Academia.edu uses cookies to personalize content, tailor ads and improve the user experience. Find the values of the term for f(x+h) and f(x) by identifying x and h. Simplify the expression under the limit and cancel common factors whenever possible. %%EOF Being the first major exam in your life, preparing for them can be very challenging. This is also known as the first derivative of the function. /Length 1836 Problem-solving Draw a sketch showing points A and B and the chord between them. To derive the differentiation of the trigonometric function sin x, we will use the following limit and trigonometric formulas: sin (A+B) = sin A cos B + sin B cos A limx0 cosx1 x = 0 lim x 0 cos x 1 x = 0 Learnderivatives of cos x,derivatives of sin x,derivatives of xsinxandderivative of 2xhere. Derivative by the first principle refers to using algebra to find a general expression for the slope of a curve. The only trick needed is that you reduce the power of each term by one, and put an 'n' in front. For example, the gradient of the below curve at A is equal to the gradient of the tangent at A, which . In this article, we will learn to find the rate of change of one variable with respect to another variable using the First Principle of Differentiation. (a) Given that , find from first principles. Answer: d dx sinx = cosx Explanation: By definition of the derivative: f '(x) = lim h0 f (x + h) f (x) h So with f (x) = sinx we have; f '(x) = lim h0 sin(x +h) sinx h %PDF-1.5 % Over two thousand years ago, Aristotle defined a first principle as "the first basis from which a thing is known." First principles thinking is a fancy way of saying "think like a scientist." Scientists don't assume anything. A curve does not have a constant gradient. Step 3: After that click on that link than automatically the PDF will be downloaded. Calculus Differentiating Trigonometric Functions Differentiating sin (x) from First Principles Key Questions How do you differentiate f (x) = sin(x) from first principles? Figure 2. pdf, 393.23 KB pptx, 1.24 MB " Differentiation from First Principles " starts by illustrating the difficulty of finding a reliable answer for the gradient at a point on a curve by drawing a tangent to the curve. the first principles approach above if you are asked to. To learn more, view ourPrivacy Policy. Prove from first principles that the derivative of x3 is 3x2 (5) 2. 224 0 obj <>/Filter/FlateDecode/ID[<474B503CD9FE8C48A9ACE05CA21A162D>]/Index[202 43]/Info 201 0 R/Length 103/Prev 127199/Root 203 0 R/Size 245/Type/XRef/W[1 2 1]>>stream They are also useful to find Definite Integral by Parts, Exponential Function, Trigonometric Functions, etc. Formula for First principle of Derivatives: f ( x ) = lim h 0 (f ( x + h ) f ( x )) /h. In marketing literature, differentiation refers to a strategy devised to outperform rival brands/products by providing unique features or services to make the product/brand desirable and foster . Evaluate the resulting expressions limit as h0. Integration is covered in tutorial 1. Its a crucial idea with a wide range of applications: in everyday life, the derivative can inform us how fast we are driving or assist us in predicting the stock market changes. CBSE invites ideas from teachers and students to improve education, 5 differences between R.D. Is it ok to start solving H C Verma part 2 without being through part 1? Therefore, due to one unit change in $x$, the corresponding change in $y$ is $\frac{{\Delta y}}{{\Delta x}}$. engineering. 82 - MME - A Level Maths - Pure - Differentiation from First Principles A Level Finding Derivatives from First Principles The derivatives are used to find solutions to differential equations. In this article, we are going to learn about Derivative by the first principle, the definition of the first principle of derivative, Proof of the first principle of derivative, One-sided derivative, Derivatives of trigonometric functions using the first principle, the derivative of sinx, cosx and tanx by the first principle with solved examples and FAQs, The derivative of a function is a concept in mathematicsof real variable that measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). If the one-sided derivatives are equal, then the function has an ordinary derivative at x_o. Write down the formula for finding the derivative using first principles g ( x) = lim h 0 g ( x + h) g ( x) h Determine g ( x + h) We also learnt that the derivative of a function $y = f(x)$ at a point is the slope of the tangent to the curve at that point. Hope this article on the First Principles of Derivatives was informative. - Practice questions with answers, including interactive assesment code. The left-hand derivative and right-hand derivative are defined by: $\begin{matrix} f_{-}(a)=\lim _{h{\rightarrow}{0^-}}{f(a+h)f(a)\over{h}}\\ f_{+}(a)=\lim _{h{\rightarrow}{0^+}}{f(a+h)f(a)\over{h}} \end{matrix}$. First principles thinking consists of deriving things to their fundamental proven axioms in the given arena, before reasoning up by asking which ones are relevant to the question at hand, then cross referencing conclusions based on chosen axioms and making sure conclusions don't violate any fundamental laws. (5) 3. Differentiation From First Principles The aim of differentiation is to find the gradient of the tangent lines to a curve. Write down the formula for finding the derivative using first principles g ( x) = lim h 0 g ( x + h) g ( x) h Determine g ( x + h) First Principles of Differential Calculus Differentiation is about finding the instantaneous rate of change of a function. Explain how the answer to part (i) relates to the gradient of the curve at A. Already have an account? First Principles of Derivatives are useful for finding Derivatives of Algebraic Functions, Derivatives of Trigonometric Functions, Derivatives of Logarithmic Functions. The First Principle of Differentiation We will now derive and understand the concept of the first principle of a derivative. By taking two points on the curve that lie very closely together, the straight line between them will have approximately the same gradient as the tangent there. It is the instantaneous rate of change of a function at a point in its domain. >> Example Consider the straight line y = 3x+2 shown in Figure 1. Differentiate $\sqrt {4 x} $ with respect to $x$ from first principle.Ans: Given: $f(x) = \sqrt {4 x} $${f^\prime }(x) = \mathop {\lim }\limits_{h \to 0} \frac{{f(x + h) f(x)}}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\sqrt {4 (x + h)} \sqrt {4 x} }}{h}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{\left[ {\sqrt {4 (x + h)} \sqrt {4 x} } \right]\left[ {\sqrt {4 (x + h)} + \sqrt {4 x} } \right]}}{{h\left[ {\sqrt {4 (x + h)} + \sqrt {4 x} } \right]}}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{4 (x + h) (4 x)}}{{h\left[ {\sqrt {4 (x + h)} + \sqrt {4 x} } \right]}}$$ = \mathop {\lim }\limits_{h \to 0} \frac{{ h}}{{h\left[ {\sqrt {4 x h} + \sqrt {4 x} } \right]}}$$\therefore \frac{d}{{dx}}(\sqrt {4 x} ) = \frac{{ 1}}{{2\sqrt {4 x} }}$, Q.3. If you look at the graph of (x) = x/2 (below), you can see that when x increases by two ( 2 ), y increases by one ( 1 ). Differentiate from first principles y = 2x2 (5) A-Level Pt. -Sl-sk -3 [51 S +k) 43 (b) Given that y = + 2x2 and www.naikermaths.com = 7 when x = 4, find the value of the constant a. If it doesn't start automatically than save it manually in the drive. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. Calculus is usually divided up into two parts, integration and differentiation. In this unit we look at how to differentiate very simple functions from first principles. Perhaps this is the point of confusion. xZo8~_{KF[rvmiKmd[Nd'^H)eF?N/ T-d!Bv%+a uK^'&RhN1&c(dv64E(fwX"2 tKv1MZU11QmQ]mFr.V"8'V6@$5JiS=:VCU Differentiation from first principles of some simple curves For any curve it is clear that if we choose two points and join them, this produces a straight line. [41 The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. DN 1.1: Differentiation from First Principles Page 2 of 3 June 2012 2. \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=sinx\\ f(x+h)=sin(x+h)\\ f(x+h)f(x)= sin(x+h) sin(x) = sinxcosh + cosxsinh sinx\\ = sinx(cosh-1) + cosxsinh\\ {f(x+h) f(x)\over{h}}={ sinx(cosh-1) + cosxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sinx(cosh-1) + cosxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinx(cosh-1)\over{h}} + \lim _{h{\rightarrow}0} {cosxsinh\over{h}}\\ = sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} + cosx \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \text{Put h = 0 in first limit}\\ sinx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = sinx\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \lim _{h{\rightarrow}0} {cosh\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cosx \times1 = cosx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx \end{matrix}\), $\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=sinx\\ f(x+h)=sin(x+h)\\ f(x+h)f(x)= sin(x+h) sin(x) = {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {2cos({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0}2cos({x+h+x\over{2}}){sin({x+h-x\over{2}})\over{{h\over{2}}}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}})\times1\\ {\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} 2cos({x+h+x\over{2}}) = cosx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = cosx \end{matrix}$, \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=cosx\\ f(x+h)=cos(x+h)\\ f(x+h)f(x)= cos(x+h) cos(x) = cosxcosh sinxsinh cosx\\ = cosx(cosh-1) sinxsinh\\ {f(x+h) f(x)\over{h}}={ cosx(cosh-1) sinxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { cosx(cosh-1) sinxsinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosx(cosh-1)\over{h}} \lim _{h{\rightarrow}0} {sinxsinh\over{h}}\\ = cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} sinx \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \text{Put h = 0 in first limit}\\ cosx \lim _{h{\rightarrow}0} {(cosh-1)\over{h}} = cosx\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {{d\over{dh}}sinh\over{{d\over{dh}}h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \lim _{h{\rightarrow}0} {cosh\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = -sinx \times1 = -sinx\\ f(x)={dy\over{dx}} = {d(cosx)\over{dx}} = -sinx \end{matrix}\), $\begin{matrix}\ f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=cosx\\ f(x+h)=cos(x+h)\\ f(x+h)f(x)= cos(x+h) cos(x) = {-2sin({x+h+x\over{2}})sin({x+h-x\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {-2sin({2x+h\over{2}})sin({h\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2cos(x+{h\over{2}}){sin({h\over{2}})\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0}-2sin(x+{h\over{2}}){sin({h\over{2}})\over{{h\over{2}}}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}})\times1\\ {\because}\lim _{h{\rightarrow}0}{sin({h\over{2}})\over{{h\over{2}}}} = 1\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} -2sin(x+{h\over{2}}) = -sinx\\ f(x)={dy\over{dx}} = {d(sinx)\over{dx}} = -sinx \end{matrix}$, If f(x) = tanx , find f(x) \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}} f(x)=tanx\\ f(x+h)=tan(x+h)\\ f(x+h)f(x)= tan(x+h) tan(x) = {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\\ {f(x+h) f(x)\over{h}}={ {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { {sin(x+h)\over{cos(x+h)}} {sin(x)\over{cos(x)}}\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {cosxsin(x+h) sinxcos(x+h)\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {{sin(2x+h)+sinh\over{2}} {sin(2x+h)-sinh\over{2}}\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{hcosxcos(x+h)}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sinh\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {1\over{cosxcos(x+h)}}\\ =1\times{1\over{cosx\times{cosx}}}\\ ={1\over{cos^2x}}\\ ={sec^2x}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = {sec^2x}\\ f(x)={dy\over{dx}} = {d(tanx)\over{dx}} = {sec^2x} \end{matrix}\), \(\begin{matrix} f(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)f(x)\over{h}}\\ f(x)=sin5x\\ f(x+h)=sin(5x+5h)\\ f(x+h)f(x)= sin(5x+5h) sin(5x) = sin5xcos5h + cos5xsin5h sin5x\\ = sin5x(cos5h-1) + cos5xsin5h\\ {f(x+h) f(x)\over{h}}={ sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} { sin5x(cos5h-1) + cos5xsin5h\over{h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = \lim _{h{\rightarrow}0} {sin5x(cos5h-1)\over{h}} + \lim _{h{\rightarrow}0} {cos5xsin5h\over{h}}\\ = sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} + cos5x \lim _{h{\rightarrow}0} {sin5h\over{h}}\\ \text{Put h = 0 in first limit}\\ sin5x \lim _{h{\rightarrow}0} {(cos5h-1)\over{h}} = sin5x\times0 = 0\\ \text{Using L Hospitals Rule on Second Limit}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} 5\times{{d\over{dh}}sin5h\over{{d\over{dh}}5h}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \lim _{h{\rightarrow}0} {5cos5h\over{1}}\\ \lim _{h{\rightarrow}0}{f(x+h) f(x)\over{h}} = cos5x \times5 = 5cos5x \end{matrix}\). 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