Then is differentiable at if and only if there exists an by matrix such that the "error" function has the property that approaches as approaches . Find the derivative of the function \begin{equation} y=\frac{x}{\sqrt{x^4+4}}. PQk< , then kf(Q) f(P)k0 such that if k! Suppose that $u=g(x)$ is differentiable at $x=-5,$ $y=f(u)$ is differentiable at $u=g(-5),$ and $(f\circ g)'(-5)$ is negative. This speculation turns out to be correct, but we would like a better justification that what is perhaps a happenstance of notation. If Δx is an increment in x and Δu and Δy are the corresponding increment in u and y, then we can use Equation(1) to write Δu = g’(a) Δx + ε 1 Δx = * g’(a) + ε x��YK�5��W7�`�ޏP�@ Dave4Math » Calculus 1 » The Chain Rule (Examples and Proof). Proof of the Chain Rule •Suppose u = g(x) is differentiable at a and y = f(u) is differentiable at b = g(a). (a) Find the tangent to the curve $y=2 \tan (\pi x/4)$ at $x=1.$ (b) What is the smallest value the slope of the curve can ever have on the interval $-2> Example. By the chain rule $$ g'(x)=f'(3x-1)\frac{d}{dx}(3x-1)=3f'(3x-1)=\frac{3}{(3x-1)^2+1}. On the other hand, simple basic functions such as the fifth root of twice an input does not fall under these techniques. Using the chain rule and the quotient rule, \begin{equation} \frac{dy}{dx}=\frac{\sqrt{x^4+4}(1)-x\frac{d}{dx}\left(\sqrt{x^4+4}\right)}{\left(\sqrt{x^4+4}\right)^2}=\frac{\sqrt{x^4+4}(1)-x\left(\frac{2 x^3}{\sqrt{4+x^4}}\right)}{\left(\sqrt{x^4+4}\right)^2} \end{equation} which simplifies to \begin{equation} \frac{dy}{dx}=\frac{4-x^4}{\left(4+x^4\right)^{3/2}} \end{equation} as desired. Show that \begin{equation} \frac{d}{d x}( \ln |\cos x| )=-\tan x \qquad \text{and}\qquad \frac{d}{d x}(\ln|\sec x+\tan x|)=\sec x. Example. 2 Chain rule for two sets of independent variables If u = u(x,y) and the two independent variables x,y are each a function of two new independent variables s,tthen we want relations between their partial derivatives. We wish to show $ \frac{d f}{d x}=\frac{df}{du}\frac{du}{dx}$ and will do so by using the definition of the derivative for the function $f$ with respect to $x,$ namely, \begin{equation} \frac{df}{dx}=\lim_{\Delta x\to 0}\frac{f[u(x+\Delta x)]-f[u(x)]}{\Delta x} \end{equation} To better work with this limit let’s define an auxiliary function: \begin{equation} g(t)= \begin{cases} \displaystyle \frac{f[u(x)+t]-f[u(x)]}{t}-\frac{df}{du} & \text{ if } t\neq 0 \\ 0 & \text{ if } t=0 \end{cases} \end{equation} Let $\Delta u=u(x+\Delta x)-u(x),$ then three properties of the function $g$ are. Practice: Chain rule capstone. To begin with, let us introduce a variable u = g(x) to simplify the looks of our steps. let t = 1 + x² therefore, y = t³ dy/dt = 3t² dt/dx = 2x by the Chain Rule, dy/dx = dy/dt × dt/dx so dy/dx = 3t² × 2x = 3(1 + x²)² × 2x = 6x(1 + x²)² Batool Akmal. This section gives plenty of examples of the use of the chain rule as well as an easily understandable proof of the chain rule. It is especially transparent using o() notation, where once again f(x) = o(g(x)) means that lim x!0 f(x) g(x) = 0: Also related to the tangent approximation formula is the gradient of a function. Using the chain rule, \begin{align} \frac{dy}{dx}&=\cos \sqrt[3]{x}\frac{d}{dx}\left(\sqrt[3]{x}\right)+\frac{1}{3}(\sin x)^{-2/3}\frac{d}{dx}(\sin x) \\ & =\frac{1}{3 x^{2/3}}\cos \sqrt[3]{x}+\frac{\cos x}{3(\sin x)^{2/3}}. �Vq ���N�k?H���Z��^y�l6PpYk4ږ�����=_^�>�F�Jh����n� �碲O�_�?�W�Z��j"�793^�_=�����W��������b>���{� =����aޚ(�7.\��� l�����毉t�9ɕ�n"�� ͬ���ny�m�`�M+��eIǬѭ���n����t9+���l�����]��v���hΌ��Ji6I�Y)H\���f Using the chain rule and the product rule we determine, \begin{equation} g'(x)=2x f\left(\frac{x}{x-1}\right)+x^2f’\left(\frac{x}{x-1}\right)\frac{d}{dx}\left(\frac{x}{x-1}\right)\end{equation} \begin{equation} = 2x f\left(\frac{x}{x-1}\right)+x^2f’\left(\frac{x}{x-1}\right)\left(\frac{-1}{(x-1)^2}\right). \end{equation}. $$ Also, by the chain rule \begin{align} h'(x) & = f’\left(\frac{1}{x}\right)\frac{d}{dx}\left(\frac{1}{x}\right) \\ & =-f’\left(\frac{1}{x}\right)\left(\frac{1}{x^2}\right) \\ & =\frac{-1}{\left(\frac{1}{x} \right)^2 + 1} \left(\frac{1}{x^2}\right) \\ & =\frac{-1}{x^2+1}. Let $f$ be a function for which $$ f'(x)=\frac{1}{x^2+1}. This proof uses the following fact: Assume , and . $$ as desired. Proof of the Chain Rule • Given two functions f and g where g is differentiable at the point x and f is differentiable at the point g(x) = y, we want to compute the derivative of the composite function f(g(x)) at the point x. Proof. The chain rule is used for linking parts of equations together or for differentiating complicated equations like nested equations. It's a "rigorized" version of the intuitive argument given above. Using the differentiation rule $\frac{d}{dx}[\ln u]=\frac{u’}{u};$ we have, \begin{equation} \frac{d}{d x}( \ln |\cos x| ) =\frac{1}{\cos x}\frac{d}{dx}(\cos x) =\frac{\sin x}{\cos x} =\tan x \end{equation} and \begin{align} & \frac{d}{d x}( (\ln |\sec x+\tan x|) ) \\ & \qquad =\frac{1}{|\sec x+\tan x|}\frac{d}{dx}(|\sec x+\tan x|) \\ & \qquad = \frac{1}{|\sec x+\tan x|}\frac{\sec x+\tan x}{|\sec x+\tan x|}\frac{d}{dx}(\sec x+\tan x) \\ & \qquad =\frac{1}{|\sec x+\tan x|}\frac{\sec x+\tan x}{|\sec x+\tan x|}(\sec x \tan x +\sec^2 x)\\ & \qquad =\frac{\sec x \tan x+\sec ^2x}{\sec x+\tan x} \\ & \qquad =\sec x \end{align} using $\displaystyle \frac{d}{dx}[|u|]=\frac{u}{|u|}(u’), u\neq 0.$, Example. When u = u(x,y), for guidance in working out the chain rule, write down the differential δu= ∂u ∂x δx+ ∂u ∂y δy+ ... (3) then when x= x(s,t) and y= y(s,t) (which are known functions of sand t), the … Given $y=6u-9$ and find $\frac{dy}{dx}$ for (a) $u=(1/2)x^4$, (b) $u=-x/3$, and (c) $u=10x-5.$, Exercise. Then the previous expression is equal to the product of two factors: Find the derivative of the function \begin{equation} y=\sin \sqrt[3]{x}+\sqrt[3]{\sin x} \end{equation}, Solution. Purported Proof of the Chain Rule: Recall that dy du = f0(u) = lim ∆u→0 f(u+∆u)−f(u) ∆u and let u = g(x) and ∆u = ∆g = g(x+∆x)−g(x). The chain rule can be used iteratively to calculate the joint probability of any no.of events. In Calculus, a Quotient rule is similar to the product rule. In single-variable calculus, we found that one of the most useful differentiation rules is the chain rule, which allows us to find the derivative of the composition of two functions. The chain rule is used to differentiate composite functions. Solution. First proof. Given: Functions and . dy/dx = F'(H(x)).H'(x) dy/dx = F'(H(x)).H(x) dy/dx = F'(H(x)) dy/dx = F'(H(x)) / H'(x) dy/dx = F'(H(x)) + H'(x) Author of lecture Chain Rule Proof. Derivative rules review. Specifically, it allows us to use differentiation rules on more complicated functions by differentiating the inner function and outer function separately. It follows that f0[g(x)] = lim ∆g→0 f[g(x)+∆g]−f[g(x)] ∆g = lim ∆x→0 f[g(x+∆x)]−f[g(x)] g(x+∆x)−g(x) = lim ∆x→0 f [ g ( x)] – f [ g ( c)] x – c = Q [ g ( x)] g ( x) − g ( c) x − c. for all x in a punctured neighborhood of c. In which case, the proof of Chain Rule can be finalized in a few steps through the use of limit laws. Note that now in terms of this new variable, we need to prove: ddxf(u)=f′(u)dudx{\frac{d}{dx}f(u) = f'(u)\frac{du}{dx}} dxd​f(u)=f′(u)dxdu​ Since u = g(x), it is obviously a function of x; and since we have assumed g(x) to be differentiable, u will also be differentiable. Lecture 3: Chain Rules and Inequalities Last lecture: entropy and mutual information This time { Chain rules { Jensen’s inequality { Log-sum inequality { Concavity of entropy { Convex/concavity of mutual information Dr. Yao Xie, ECE587, Information Theory, Duke University Here we sketch a proof of the Chain Rule that may be a little simpler than the proof presented above. Theorem. $$ Also, by the chain rule \begin{align} h'(x) & = f’\left(\frac{1}{x}\right)\frac{d}{dx}\left(\frac{1}{x}\right) \\ & =-f’\left(\frac{1}{x}\right)\left(\frac{1}{x^2}\right) \\ & =\frac{-1}{\left(\frac{1}{x} \right)^2 + 1} \left(\frac{1}{x^2}\right) \\ & =\frac{-1}{x^2+1}. When will these derivatives be the same? By the chain rule, \begin{equation} a(t)=\frac{dv}{dt}=\frac{dv}{d s}\frac{ds}{dt}=v(t)\frac{dv}{ds} \end{equation} In the case where $s(t)=-2t^3+4t^2+t-3; $ we determine, \begin{equation} \frac{ds}{dt} = v(t) = -6t^2+8t+1 \qquad \text{and } \qquad a(t)=-12t+8. We now turn to a proof of the chain rule. Implicit differentiation. =_.���tK���L���d�&-.Y�Y&M6���)j-9Ә��cA�a�h,��4���2�e�He���9Ƶ�+nO���^b��j�(���{� If $y$ is a differentiable function of $u,$ $u$ is a differentiable function of $v,$ and $v$ is a differentiable function of $x,$ then $$ \frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}. The standard proof of the multi-dimensional chain rule can be thought of in this way. By the chain rule $$ g'(x)=f'(3x-1)\frac{d}{dx}(3x-1)=3f'(3x-1)=\frac{3}{(3x-1)^2+1}. $$, Exercise. And with that, we’ll close our little discussion on the theory of Chain Rule as of now. /Filter /FlateDecode Find the derivative of the function \begin{equation} y=\sin ^4\left(x^2-3\right)-\tan ^2\left(x^2-3\right). Solution. PQk: Proof. 1. A Quotient Rule is stated as the ratio of the quantity of the denominator times the derivative of the numerator function … \end{align} as desired. It is used where the function is within another function. Copyright © 2020 Dave4Math LLC. As in single variable calculus, there is a multivariable chain rule. $$ If $\displaystyle g(x)=x^2f\left(\frac{x}{x-1}\right),$ what is $g'(2)?$. Example. The Chain Rule - a More Formal Approach Suggested Prerequesites: The definition of the derivative, The chain rule. In fact, the chain rule says that the first rate of change is the product of the other two. The inner function is the one inside the parentheses: x 2 -3. V Exercise. $$ Now we can rewrite $\displaystyle \frac{df}{dx}$ as follows: \begin{align} \frac{df}{dx} & = \lim_{\Delta x\to 0}\frac{f[u(x+\Delta x)]-f[u(x)]}{\Delta x} \\ & =\lim_{\Delta x\to 0}\frac{f[u(x)+\Delta u]-f[u(x)]}{\Delta x} \\ & =\lim_{\Delta x\to 0} \frac{\left(g(\Delta u)+\frac{df}{du}\right)\Delta u}{\Delta x} \\ & =\lim_{\Delta x\to 0}\left(g(\Delta u)+\frac{df}{du}\right)\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} \\ & =\left[\lim_{\Delta x\to 0}g(\Delta u)+\lim_{\Delta x\to 0}\frac{df}{du}\right]\text{ }\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} \\ & =\left[g\left( \lim_{\Delta x\to 0}\Delta u \right)+\lim_{\Delta x\to 0}\frac{df}{du}\right]\text{ }\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} \\ & =\left[g(0)+\lim_{\Delta x\to 0}\frac{df}{du}\right]\text{ }\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} \\ & =\left[0+\lim_{\Delta x\to 0}\frac{df}{du}\right]\text{ }\lim_{\Delta x\to 0}\frac{\Delta u}{\Delta x} \\ & =\frac{df}{du}\frac{du}{dx}. By using the chain rule we determine, \begin{equation} f'(x)=\frac{2}{3}\left(9-x^2\right)^{-1/3}(-2x)=\frac{-4x}{3\sqrt[3]{9-x^2}} \end{equation} and so $\displaystyle f'(1)=\frac{-4}{3\sqrt[3]{9-1^2}}=\frac{-2}{3}.$ Therefore, an equation of the tangent line is $y-4=\left(\frac{-2}{3}\right)(x-1)$ which simplifies to $$ y=\frac{-2}{3}x+\frac{14}{3}. Assuming that the following derivatives exists, find \begin{equation} \frac{d}{d x}f’ [f(x)] \qquad \text{and}\qquad \frac{d}{d x}f [f'(x)]. $(1) \quad \displaystyle g(\Delta u)=\frac{f[u(x)+\Delta u]-f[u(x)]}{\Delta u}-\frac{df}{du}$ provided $\Delta u\neq 0$ $(2) \quad \displaystyle \left[g(\Delta u)+\frac{df}{du}\right]\Delta u=f[u(x)+\Delta u]-f[u(x)]$ $(3) \quad g$ is continuous at $t=0$ since $$ \lim_{t\to 0} \left[ \frac{f[u(x)+t]-f[u(x)]}{t}\right]=\frac{df}{du}. and M.S. Determine if the following statement is true or false. Solution. The lecture Chain Rule Proof by Batool Akmal is from the course Quotient Rule, Chain Rule and Product Rule. \end{equation}. Rm be a function. Find the derivative of the function \begin{equation} h(t)=2 \cot ^2(\pi t+2). To prove: wherever the right side makes sense. ), Calculus (Start Here) – Enter the World of Calculus, Continuous (It’s Meaning and Applications), Derivative Definition (The Derivative as a Function), Derivative Examples (The Role of the Derivative), Find the Limit (Techniques for Finding Limits), First Derivative Test (and Curve Sketching), Horizontal Asymptotes and Vertical Asymptotes, Implicit Differentiation (and Logarithmic Differentiation), L ‘Hopital’s Rule and Indeterminate Forms, Limit Definition (Precise Definition of Limit), Choose your video style (lightboard, screencast, or markerboard). Many undergraduate calculus texts assert the following proof of the chain rule. \end{equation} Therefore, \begin{equation} g'(2)=2(2) f\left(\frac{2}{2-1}\right)+2^2f’\left(\frac{2}{2-1}\right)\left(\frac{-1}{(2-1)^2}\right)=-24. Solution. And, if you've been following some of the videos on "differentiability implies continuity", and what happens to a continuous function as our change in x, if x is our independent variable, as that approaches zero, how the change in our function approaches zero, then this proof is actually surprisingly straightforward, so let's just get to it, and this is just one of many proofs of the chain rule. $$. Example. Then justify your claim. \end{align} as needed. \end{equation}, Proof. Example. Example. The single-variable chain rule. In order to understand the chin rule the reader must be aware of composition of functions. You can use our resources like sample question papers and Maths previous years’ papers to practise questions and answers for Maths board exam preparation. In other words, we want to compute lim. Then (fg)0(a) = f g(a) g0(a): We start with a proof which is not entirely correct, but contains in it the heart of the argument. Next lesson. In differential calculus, the chain rule is a way of finding the derivative of a function. Example. Using the chain rule and the quotient rule, we determine, \begin{equation} \frac{dg}{dx} =3\left(\frac{3x^2-2}{2x+3} \right)^2\left(\frac{(2x+3)6x-\left(3x^2-2\right)2}{(2x+3)^2}\right) \end{equation} which simplifies to \begin{equation} \frac{dg}{dx}=\frac{6 \left(2-3 x^2\right)^2 \left(2+9 x+3 x^2\right)}{(3+2 x)^4} \end{equation} as desired. Exercise. Solution. \end{equation} as desired. What, if anything, can be said about the values of $g'(-5)$ and $f'(g(-5))?$, Exercise. Determine the point(s) at which the graph of \begin{equation} f(x)=\frac{x}{\sqrt{2x-1}} \end{equation} has a horizontal tangent. Solution. Let's … \end{equation} Thus, \begin{equation} \frac{dv}{d s}=\frac{-12t+8}{-6t^2+8t+1}. The right side becomes: This simplifies to: Plug back the expressions and get: Your goal is to compute its derivative at a point \(t\in \R\). This field is for validation purposes and should be left unchanged. \end{equation}. If fis di erentiable at P, then there is a constant M 0 and >0 such that if k! Evaluating Limits Analytically (Using Limit Theorems) [Video], Intuitive Introduction to Limits (The Behavior of a Function) [Video], Related Rates (Applying Implicit Differentiation), Numerical Integration (Trapezoidal and Simpson’s), Integral Definition (The Definite Integral), Indefinite Integrals (What is an antiderivative? Derivative rules review. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. Let $u$ be a differentiable function of $x.$ Use $|u|=\sqrt{u^2}$ to prove that $$\frac{d}{dx}(|u| )=\frac{u’ u}{|u|} $$ when $u\neq 0.$ Use the formula to find $h’$ given $h(x)=x|2x-1|.$. The following is a proof of the multi-variable Chain Rule. Differentiate the functions given by the following equations $(1) \quad y=\cos^2\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)$$(2) \quad y=\sqrt{1+\tan \left(x+\frac{1}{x}\right)} $$(3) \quad n=\left(y+\sqrt[3]{y+\sqrt{2y-9}}\right)^8$, Exercise. Suppose that the functions $f$, $g$, and their derivatives with respect to $x$ have the following values at $x=0$ and $x=1.$ \begin{equation} \begin{array}{c|cccc} x & f(x) & g(x) & f'(x) & g'(x) \\ \hline 0 & 1 & 1 & 5 & 1/3 \\ 1 & 3 & -4 & -1/3 & -8/3 \end{array} \end{equation} Find the derivatives with respect to $x$ of the following combinations at a given value of $x,$ $(1) \quad \displaystyle 5 f(x)-g(x), x=1$ $(2) \quad \displaystyle f(x)g^3(x), x=0$ $(3) \quad \displaystyle \frac{f(x)}{g(x)+1}, x=1$$(4) \quad \displaystyle f(g(x)), x=0$ $(5) \quad \displaystyle g(f(x)), x=0$ $(6) \quad \displaystyle \left(x^{11}+f(x)\right)^{-2}, x=1$$(7) \quad \displaystyle f(x+g(x)), x=0$$(8) \quad \displaystyle f(x g(x)), x=0$$(9) \quad \displaystyle f^3(x)g(x), x=0$. Only the proof differs slightly, as the definition of the derivative is not the same. Chain rule is a formula which is the same in standard and non-standard analysis. Here is the chain rule again, still in the prime notation of Lagrange. The proof of this theorem uses the definition of differentiability of a function of two variables. Exercise. r��dͧ͜y����e,�6[&zs�oOcE���v"��cx��{���]O��� Proof. With a lot of work, we can sometimes find derivatives without using the chain rule either by expanding a polynomial, by using another differentiation rule, or maybe by using a trigonometric identity. It is used where the function is within another function. A new subsection, called "Proof in non-standard analysis", of the section "Proofs" could be added. Define \(\phi = f\circ \mathbf g\). \end{align} as desired. $$ Thus the only point where $f$ has a horizontal tangent line is $(1,1).$, Exercise. We will need: Lemma 12.4. \end{align}, Example. The gradient is one of the key concepts in multivariable calculus. Under certain conditions, such as differentiability, the result is fantastic, but you should practice using it. $$ If $g(x)=f(3x-1),$ what is $g'(x)?$ Also, if $ h(x)=f\left(\frac{1}{x}\right),$ what is $h'(x)?$. Chain Rule If f(x) and g(x) are both differentiable functions and we define F(x) = (f ∘ g)(x) then the derivative of F (x) is F ′ (x) = f ′ (g(x)) g ′ (x). \end{align} as desired. Example. The following chain rule examples show you how to differentiate (find the derivative of) many functions that have an “ inner function ” and an “ outer function.” For an example, take the function y = √ (x 2 – 3). By using the chain rule we determine, \begin{align} f'(x) & = \frac{\sqrt{2x-1}(1)-x\frac{d}{dx}\left(\sqrt{2x-1}\right)}{\left(\sqrt{2x-1}\right)^2} \\ & =\frac{\sqrt{2x-1}(1)-x \left(\frac{1}{\sqrt{-1+2 x}}\right)}{\left(\sqrt{2x-1}\right)^2} \end{align} which simplifies to $$ f'(x)=\frac{-1+x}{(-1+2 x)^{3/2}}. What is the gradient of y = F(H(x)) according to the chain rule? Worked example: Derivative of ∜(x³+4x²+7) using the chain rule. %PDF-1.4 Sort by: Top Voted. Proof: Consider the function: Its partial derivatives are: Define: By the chain rule for partial differentiation, we have: The left side is . Let AˆRn be an open subset and let f: A! This is the currently selected item. Proving the chain rule. Included Quiz Questions. f (z) = √z g(z) = 5z −8 f ( z) = z g ( z) = 5 z − 8. then we can write the function as a composition. 5 Stars: 5: … Given a2R and functions fand gsuch that gis differentiable at aand fis differentiable at g(a). One proof of the chain rule begins with the definition of the derivative: (∘) ′ = → (()) − (()) −. 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