Chain Rule

Introduction

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.

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


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