Caluclus: Table of contents

Chain Rule


Our goal is to find the derivatives of compositions of functions such as \(\cos\left(e^{x^2}+3x\sin x\right)\) and \(\cos(x^2+9)\).

Composition of functions

We will first review the composition of functions. If you want to read more on this, please follow this link Functions in Mathematics.

Assume that \(u(x)=x^3+4x\) and \(v(x)=\cos x\). Then \(v\) is a function that maps a real number \(x\) to \(\cos x\). The function \(u\) maps any real number \(x\) to \(x^3+4x\). In particular \(u(7)=7^3+4\cdot 7\), \(u(\star)=\star^3+4\cdot \star\), and \(u(v(x))=(v(x))^3+4\cdot v(x)= \cos^3x+4\cos x\). The composition \(u(v(x))\) is another function and it maps \(x\) to \(\cos^3x+4\cos x\). If we call \(f(x)=\cos^3x+4\cos x\) we can talk about the derivative of \(f\).

Example Assume that \(f(x)=2^x\) and \(g(x)=5\cdot x-6\). Let \(h\) be the function defined as \(h(x)=f(g(x))\) and let \(m\) be the function defined as \(m(x)=g(f(x))\). Determine \(h(3)\) and \(m(3)\).

Derivative of a composition of functions

We will now derive a formula for the derivative of a composition of functions \(f(x)=u(v(x))\) that involves the derivatives of \(u\) and \(v\). This formula is known as the chain rule.

Theorem (Chain rule) Assume that \(u\) and \(v\) are two functions such that \(v\) is differentiable at point \(a\), and \(u\) is differentiable at point \(v(a)\). Then the function \(f(x)=u(v(x))\) is differentiable at \(a\) and its derivative satisfies: \[f^{\prime}(a)=u^{\prime}(v(a))\cdot v^{\prime}(a).\]

Let us consider the following example.

Example Prove that the function \(f(x)=\cos(3x+x^2)\) is differentiable and find its derivative \(f^{\prime}(x)\) for \(x\in\mathbb R\).

Derivative of the inverse function

We will derive a formula that will help us in finding derivatives of \(\ln\) and inverse trigonometric functions.

Theorem (Derivative of inverse function) Let \(f:\mathbb R\to\mathbb R\) be a continuously differentiable function at \(a\in\mathbb R\) for which \(f^{\prime}(a)\neq 0\). Then its inverse \(g(z)=f^{-1}(z)\) is differentiable at \(b=f^{-1}(a)\) and its derivative satisfies \[g^{\prime}(b)=\frac1{f^{\prime}\left(f^{-1}(b)\right)}.\]

We can now find the derivative of \(\ln\), \(\arcsin\), \(\arccos\), and \(\arctan\).

Let \(f(x)=e^x\). Then \(f^{-1}(x)=\ln x\) and according to the previous theorem we have \[\left(\ln x\right)^{\prime}=\frac1{f^{\prime}(\ln x)}=\frac1{e^{\ln x}}=\frac1x.\]

Let \(f(x)=\sin x\). Then \(f^{-1}(x)=\arcsin x\), hence \[\left(\arcsin x\right)^{\prime}=\frac1{f^{\prime}(\arcsin x)}=\frac1{\cos(\arcsin x)}=\frac1{\sqrt{1-x^2}},\;\; \;\mbox{for }x\in(-1,1).\]

Practice problems

Problem 1. If \(u(x)=\sin x\) and \(v(x)=x^2\), determine the \(u(v(x))\).

Problem 2. If \(u(x)= x^3\) and \(v(x)=x+4\), and \(w(x)=\cos (x+1)\), determine the \(u(v(w(x)))\).

Problem 3. Find the derivative of the function \(f(x)=\cos(x^3)\).

Problem 4. Find the derivative of the function \(f(x)=e^{-\frac{x^2}2}\).

Problem 5. Assume that \(u\) and \(v\) are the functions that satisfy: \(u(0)=1\), \(u(1)=2\), \(u(2)=3\), \(v(0)=2\), \(v(1)=3\), \(v(2)=4\), \(u^{\prime}(0)=4\), \(u^{\prime}(1)=5\), \(u^{\prime}(2)=6\), \(v^{\prime}(0)=7\), \(v^{\prime}(1)=8\), \(v^{\prime}(2)=9\). Let \(f(x)=u(v(x))\). Find \(f^{\prime}(0)\).