Proof: Consider the function: Its partial derivatives are: Define: By the chain rule for partial differentiation, we have: The left side is . 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. Thus $$\phi$$ is a function $$\R\to \R$$. A new subsection, called "Proof in non-standard analysis", of the section "Proofs" could be added. In Calculus, a Quotient rule is similar to the product rule. The inner function is the one inside the parentheses: x 2 -3. $$If \displaystyle g(x)=x^2f\left(\frac{x}{x-1}\right), what is g'(2)?. Evaluating Limits Analytically (Using Limit Theorems) [Video], Intuitive Introduction to Limits (The Behavior of a Function) [Video], Related Rates (Applying Implicit Differentiation), Numerical Integration (Trapezoidal and Simpson’s), Integral Definition (The Definite Integral), Indefinite Integrals (What is an antiderivative? Then justify your claim. If fis di erentiable at P, then there is a constant M 0 and >0 such that if k! Purported Proof of the Chain Rule: Recall that dy du = f0(u) = lim ∆u→0 f(u+∆u)−f(u) ∆u and let u = g(x) and ∆u = ∆g = g(x+∆x)−g(x). And with that, we’ll close our little discussion on the theory of Chain Rule as of now. We will need: Lemma 12.4. We now turn to a proof of the chain rule. Suppose f is a differentiable function on \mathbb{R}. Let F and G be the functions defined by$$ F(x)=f(\cos x) \qquad \qquad G(x)=\cos (f(x)). This section gives plenty of examples of the use of the chain rule as well as an easily understandable proof of the chain rule. �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 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. Translating the chain rule into Leibniz notation. 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 r��dͧ͜y����e,�6[&zs�oOcE���v"��cx��{���]O��� Then the previous expression is equal to the product of two factors: \end{align} as desired. Show that if a particle moves along a straight line with position $s(t)$ and velocity $v(t),$ then its acceleration satisfies $a(t)=v(t)\frac{dv}{ds}.$ Use this formula to find $\frac{dv}{d s}$ in the case where $s(t)=-2t^3+4t^2+t-3.$. Solution. 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: What is the gradient of y = F(H(x)) according to the chain rule? Solution. /Length 1995 The Chain Rule and the Extended Power Rule section 3.7 Theorem (Chain Rule)): Suppose that the function f is ﬀtiable at a point x and that g is ﬀtiable at f(x) .Then the function g f is ﬀtiable at x and we have (g f)′(x) = g′(f(x))f′(x)g f(x) x f g(f(x)) Note: So, if the derivatives on the right-hand side of the above equality exist , then the derivative Example. In fact, the chain rule says that the first rate of change is the product of the other two. 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, $$\frac{df}{dx}=\lim_{\Delta x\to 0}\frac{f[u(x+\Delta x)]-f[u(x)]}{\Delta x}$$ To better work with this limit let’s define an auxiliary function: $$g(t)= \begin{cases} \displaystyle \frac{f[u(x)+t]-f[u(x)]}{t}-\frac{df}{du} & \text{ if } t\neq 0 \\ 0 & \text{ if } t=0 \end{cases}$$ Let $\Delta u=u(x+\Delta x)-u(x),$ then three properties of the function $g$ are. >> V Dave4Math » Calculus 1 » The Chain Rule (Examples and Proof). To simplify the set-up, let’s assume that $$\mathbf g:\R\to \R^n$$ and $$f:\R^n\to \R$$ are both functions of class $$C^1$$. The outer function is √ (x). The chain rule is used to differentiate composite functions. Using the chain rule and the product rule we determine, $$g'(x)=2x f\left(\frac{x}{x-1}\right)+x^2f’\left(\frac{x}{x-1}\right)\frac{d}{dx}\left(\frac{x}{x-1}\right)$$ $$= 2x f\left(\frac{x}{x-1}\right)+x^2f’\left(\frac{x}{x-1}\right)\left(\frac{-1}{(x-1)^2}\right). By the chain rule g'(x)=f'(3x-1)\frac{d}{dx}(3x-1)=3f'(3x-1)=\frac{3}{(3x-1)^2+1}. In the following examples we continue to illustrate the chain rule. Dave will teach you what you need to know. This speculation turns out to be correct, but we would like a better justification that what is perhaps a happenstance of notation.$$. If $y$ is a differentiable function of $u,$ $u$ is a differentiable function of $v,$ and $v$ is a differentiable function of $x,$ then $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dv}\frac{dv}{dx}. One proof of the chain rule begins with the definition of the derivative: (∘) ′ = → (()) − (()) −. Here is the chain rule again, still in the prime notation of Lagrange. Show that$$\frac{d}{d\theta }(\sin \theta {}^{\circ})=\frac{\pi }{180}\cos \theta .$$What do you think is the importance of the exercise? Exercise. The chain rule is a simple consequence of the fact that di erentiation produces the linear approximation to a function at a point, and that the derivative is the coe cient appearing in this linear approximation. Solution.$$ 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}. 1. \end{align} as desired. Under certain conditions, such as differentiability, the result is fantastic, but you should practice using it. Leibniz's differential notation leads us to consider treating derivatives as fractions, so that given a composite function y(u(x)), we guess that . Sort by: Top Voted. The following is a proof of the multi-variable Chain Rule. Assume for the moment that () does not equal () for any x near a. This is the currently selected item. The proof of this theorem uses the definition of differentiability of a function of two variables. In addition, the Maths videos and other learning resources on our study portal are of great support during … If y = (1 + x²)³ , find dy/dx . The single-variable chain rule. The derivative would be the same in either approach; however, the chain rule allows us to find derivatives that would otherwise be very difficult to handle. If $g(t)=[f(\sin t)]^2,$ where $f$ is a differentiable function, find $g'(t).$, Exercise. Suppose that f is differentiable at the point $$\displaystyle P(x_0,y_0),$$ where $$\displaystyle x_0=g(t_0)$$ and … 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 a2R and functions fand gsuch that gis differentiable at aand fis differentiable at g(a). 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. $$Find expressions for F'(x) and G'(x)., Exercise. 3 0 obj << Okay, so you know how to differentiation a function using a definition and some derivative rules. Example. Proof. Differentiate the functions given by the following equations (1) \quad y=\cos^2\left(\frac{1-\sqrt{x}}{1+\sqrt{x}}\right)$$(2) \quad y=\sqrt{1+\tan \left(x+\frac{1}{x}\right)} $$(3) \quad n=\left(y+\sqrt[3]{y+\sqrt{2y-9}}\right)^8, Exercise. in Mathematics and has enjoyed teaching precalculus, calculus, linear algebra, and number theory at both the junior college and university levels for over 20 years. . The right side becomes: This simplifies to: Plug back the expressions and get: Specifically, it allows us to use differentiation rules on more complicated functions by differentiating the inner function and outer function separately. However, we can get a better feel for it using some intuition and a couple of examples. Chain rule is a formula which is the same in standard and non-standard analysis. Find the derivative of the following functions.(1) \quad \displaystyle r=-(\sec \theta +\tan \theta )^{-1}$$(2) \quad \displaystyle y=\frac{1}{x}\sin ^{-5}x-\frac{x}{3}\cos ^3x$$(3) \quad \displaystyle y=(4x+3)^4(x+1)^{-3}$$(4) \quad \displaystyle y=(1+2x)e^{-2x}$$(5) \quad \displaystyle h(x)=x \tan \left(2 \sqrt{x}\right)+7$$(6) \quad \displaystyle g(t)=\left(\frac{1+\cos t}{\sin t}\right)^{-1}$$(7) \quad \displaystyle q=\sin \left(\frac{t}{\sqrt{t+1}}\right)$$(8) \quad \displaystyle y=\theta ^3e^{-2\theta }\cos 5\theta $$(9) \quad \displaystyle y=(1+\cos 2t)^{-4}$$(10) \quad \displaystyle y=\left(e^{\sin (t/2)}\right)^3$$(11) \quad \displaystyle y=\left(1+\tan ^4\left(\frac{t}{12}\right)\right)^3$$(12) \quad \displaystyle y=4 \sin \left(\sqrt{1+\sqrt{t}}\right)$$(13) \quad \displaystyle y=\frac{1}{9}\cot (3x-1)$$(14) \quad \displaystyle y=\sin \left(x^2e^x\right)(15) \quad \displaystyle y=e^x \sin \left(x^2e^x\right), Exercise. Using the differentiation rule \frac{d}{dx}[\ln u]=\frac{u’}{u}; we have, $$\frac{d}{d x}( \ln |\cos x| ) =\frac{1}{\cos x}\frac{d}{dx}(\cos x) =\frac{\sin x}{\cos x} =\tan x$$ 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. 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. The Chain Rule - a More Formal Approach Suggested Prerequesites: The definition of the derivative, The chain rule. 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}. $$, Exercise. PQk< , then kf(Q) f(P) Df(P)! Practice: Chain rule capstone. Therefore, $$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. Proving the chain rule.$$. Let’s see this for the single variable case rst. Let f be a function for which f(2)=-3 and$$ f'(x)=\sqrt{x^2+5}. The gradient is one of the key concepts in multivariable calculus. Solution. The chain rule is used for linking parts of equations together or for differentiating complicated equations like nested equations. Proof. Theorem. Suppose that the functions $f$, $g$, and their derivatives with respect to $x$ have the following values at $x=0$ and $x=1.$ $$\begin{array}{c|cccc} x & f(x) & g(x) & f'(x) & g'(x) \\ \hline 0 & 1 & 1 & 5 & 1/3 \\ 1 & 3 & -4 & -1/3 & -8/3 \end{array}$$ 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$. Using the chain rule and the formula $\displaystyle \frac{d}{dx}(\cot u)=-u’\csc ^2u,$ \begin{align} \frac{dh}{dt} & =4\cot (\pi t+2)\frac{d}{dx}[\cot (\pi t+2)] \\ & =-4\pi \cot (\pi t+2)\csc ^2(\pi t+2).