The chain rule

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Summary

​The chain rule

In a nutshell

You use the chain rule when functions are concatenated. This means that one function encloses a second function. For example f(x)=u(v(x))f(x)=u(v(x)). The function v(x)v(x) is inside the function u(x)u(x).



Functions

A reminder on function notation will be useful.


A function is simply a set of instructions that are applied to the term inside the function. So a function given as u(x)u(x) tells you what to do to xx. If instead you have the same function given as u(y)u(y), the instructions are applied to yy.


Example 1

A function is given as u(x)=3x2+4x2u(x)=3x^2+4x-2. Find an expression for u(y)u(y).


All you do here is replace each instance of xx with an instance of yy. Think of the term inside the function uu as a placeholder. Hence 

u(y)=3y2+4y2\underline{u(y)=3y^2+4y-2}​​


Example 2

A function is given as u(x)=3x2+4x2u(x)=3x^2+4x-2. Another function is given as v(x)=2x1v(x)=2x-1. Find an expression for u(v(x))u(v(x)).


Recall that the term inside the definition of a function is just a placeholder. So 

u(v(x))=3v(x)2+4v(x)2u(v(x))=3v(x)^2+4v(x)-2​​


But you know that v(x)=2x1v(x)=2x-1, so this expression above can be written as

u(v(x))=3(2x1)2+4(2x1)2u(v(x))=3(2x-1)^2+4(2x-1)-2​​


This is expanded and simplified to become

u(v(x))=12x24x3\underline{u(v(x))=12x^2-4x-3}


It should be noted that you would not be expected to spot the functions u(x)u(x) and v(x)v(x) from this result.



The chain rule

The chain rule concerns itself with differentiation functions of the form f(x)=u(v(x))f(x)=u(v(x)), in other words a chain of functions. In particular, unlike with example 2, the concatenation may not be easily expanded and simplified.


PROCEDURE

1.

Identify the outer function u(x)u\left(x\right) and inner function v(x)v\left(x\right) individually.

2.

Using differentiation, find u(x)u'(x) and v(x)v'(x).​

3.

The derivative of f(x)f(x) is defined as f(x)=u(v(x))v(x)f'(x)=u'(v(x))v'(x). In words this is the derivative of uu with respect to xx evaluated at v(x)v(x), multiplied by the derivative of vv with respect to xx.


Note: There are two ways to think of u(v(x))u'(v(x)). One way is described in the procedure above: it's the derivative of uu with respect to xx, but evaluated at v(x)v(x). Alternatively it is the derivative of uu with respect to v(x)v(x)


Alternative notation

Rather than using function notation when differentiating, you can use dydx\frac{\text dy}{\text dx} notation. For y=f(u)y=f(u) where u=g(x)u=g(x)​, the chain rule can be given as

dydx=dydu×dudx\frac{\text dy}{\text dx}=\frac{\text dy}{\text du}\times\frac{\text du}{\text dx}​​

This can be viewed as a substitution: you substitute some function uu into your equation before differentiating. ​


Example 3

Differentiate y=(4x5+3x2+1)7y=(4x^5+3x^2+1)^7 with respect to xx.


Make the substitution u=4x5+3x2+1u=4x^5+3x^2+1 and rewrite the equation: y=u7y=u^7. Now find dydu\frac{\text dy}{\text du} and dudx\frac{\text du}{\text dx}:

dydu=7u6 dudx=20x4+6x\frac{\text dy}{\text du}=7u^6\\\space\\\frac{\text du}{\text dx}=20x^4+6x​​


Thus

dydx=7u6×(20x4+6x)\frac{\text dy}{\text dx}=7u^6\times(20x^4+6x)​​


Finally replace the substitution uu and simplify:

dydx=7(4x5+3x2+1)6×(20x4+6x) dydx=7(20x4+6x)(4x5+3x2+1)6\frac{\text dy}{\text dx}=7(4x^5+3x^2+1)^6\times(20x^4+6x)\\\space\\\underline{\frac{\text dy}{\text dx}=7(20x^4+6x)(4x^5+3x^2+1)^6}​​


Example 4

f(x)=(x23)3f\left(x\right)=\left(x^2-3\right)^3

Differentiate f(x)f(x) with respect to xx.


The inner function v(x)v(x) is x23x^2-3. The outer function is u(x)=x3u(x)=x^3. Hence u(x)=3x2u'(x)=3x^2 and v(x)=2xv'(x)=2x. Thus

f(x)=3(x23)2u(v(x))2xv(x)f^\prime\left(x\right)=\underbrace{{3\cdot\left(x^2-3\right)}^2}_{u'(v\left(x\right))}\cdot\underbrace{2x}_{v'(x)}

Simplify:

f(x)=6x(x23)2\underline{f'(x)={6x\left(x^2-3\right)}^2}


Example 5

Find the first derivative of the function f(x)=14x3f(x)=\frac1{4x-3}.


Identify the inner and outer functions: inner: v(x)=4x3v(x)=4x-3; outer: u(x)=1x=x1u(x)=\frac1x=x^{-1}. Thus the derivatives of these are v(x)=4v'(x)=4 and u(x)=x2=1x2u'(x)=-x^{-2}=-\frac1{x^2}. Thus, using the chain rule f(x)=u(v(x))v(x)f'(x)=u'(v(x))v'(x) gives

f(x)=1v(x)24=4(4x3)2f'(x)=-\dfrac1{v(x)^2}\cdot4=\underline{-\dfrac4{(4x-3)^2}}​​


If necessary, this could be expanded and simplified further.

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Differentiating x^n

Unit 1

Differentiating x^n

Differentiating functions with multiple terms

Unit 2

Differentiating functions with multiple terms

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The chain rule

Unit 3

The chain rule

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