### Differentiation, derivatives and Taylor approximations: Rules for differentiating functions

### The chain rule

First have a look at some examples.

g'(x)&= -3\\ \\

s'(x)&=324\,x^3-972\,x^2+972\,x-324\\ \\

f'\bigl(g(x)\bigr)\cdot g'(x)&= 4(-3x+3)^{3}\cdot (-3) \\

&= -12(-27\,x^3+81\,x^2-81\,x+27)\\

&=324\,x^3-972\,x^2+972\,x-324

\end{aligned}\] Note in this case: \[\Bigl(f\bigl(g(x)\bigr)\Bigr)'=f'\bigl(g(x)\bigr)\cdot g'(x)\]

The pattern is clear and can be generalised to the **chain rule** of differentiation.

Chain Rule If \(f\) and \(g\) are functions such that \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g(x)\), then the composite function \((f\circ g)(x)=f\bigl(g(x)\bigr)\) is differentiable at \(x\), and \[(f\circ g)'(x)=\Bigl(f\bigl(g(x)\bigr)\Bigr)'= f'\bigl(g(x)\bigr)\cdot g'(x)\]

A particular case of application of the chain rule is the calculation of the derivative of a function after some linear transformation.

If \(s(x)=f(a\cdot x+b)\), then \(s'(x)= a\cdot f'(a\cdot x+b)\)

The first difficult aspect is the identification of the composition of functions. A convenient notation helps; look at two examples below.

The function \[h(x)=(3x^2-1)^5\] can be written as composition of two functions:

\(f(u)=u^5\) and \(u=3x^2-1\)

The derivatives of these functions are:

\(\displaystyle f'(u)=\frac{\dd f}{\dd u}=5u^4\) and \(\displaystyle u'(x)=\frac{\dd u}{\dd x}=6x\).

Recall that the notation \(\displaystyle \frac{\dd f}{\dd u}\) means that you differentiate \(f\) with respect to \(u\) and then optionally substitute \(u=3x^2-1\). The notation \(\displaystyle \frac{\dd u}{\dd x}\) means that you differentiate \(u\) with respect to \(x\). Then the chain rule leads to \[h'(x)=\frac{\dd\bigl(f(u(x)\bigr)}{\dd x}=\frac{\dd f}{\dd u}\cdot \frac{\dd u}{\dd x}=5(3x^2-1)^4\cdot 6x = 30x(3x^2-1)^4\]

The function \[h(t)=\sqrt{3t^2+4}\] can be written as composition of two functions:

\(f(u)=\sqrt{u}\) and \(u=3t^2+4\)

The derivatives of these functions are:

\(\displaystyle f'(u)=\frac{\dd f}{\dd u}=\bigl(\sqrt{u}\bigr)'=\bigl( u^{\frac{1}{2}}\bigr)'=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}\) and \(\displaystyle u'(t)=\frac{\dd u}{\dd t}=6t\).

Then the chain rule leads to \[h'(t)=\frac{\dd f}{\dd u}\cdot \frac{\dd u}{\dd t}=\frac{1}{2\sqrt{3t^2+4}}\cdot 6t = \frac{3t}{\sqrt{3t^2+4}}\]

Via the chain rule you can also calculate the derivative of an inverse function when you know the derivative of the function itself.

Derivative of an inverse function For a differentiable function \(f\) with derivative \(f'\) and with corresponding inverse function \(f^{-1}\) we have: \[\frac{\dd f^{-1}}{\dd x}(x)=\frac{1}{f'\bigl(f^{-1}(x)\bigr)}\]