Caluclus: Table of contents
# Chain Rule

## Introduction

## Composition of functions

**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

**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: \[f^{\prime}(a)=u^{\prime}(v(a))\cdot v^{\prime}(a).\]
**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

**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 \[g^{\prime}(b)=\frac1{f^{\prime}\left(f^{-1}(b)\right)}.\]
## 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 \(f^{\prime}(0)\).

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

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 \(f\).

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*.

Let us consider the following example.

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

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