Solution assignment 23 Chain rule

Differentiate:

$y=\sin(\sqrt{x^2+1})$

This function requires a double chain rule as we will see (the 'quote' means: the derivative of):

$y'=\cos(\sqrt{x^2+1})\cdot[\sqrt{x^2+1}]'$

First we calculate the second factor, again using the chain rule:

$[\sqrt{x^2+1}]'=\displaystyle\frac{1}{2}(x^2+1)^{-\frac{1}{2}}.(x^2+1)'=\displaystyle\frac{x}{\sqrt{x^2+1}}$

We can substitute this result in the derivative above:

$y'=\cos(\sqrt{x^2+1})\cdot\displaystyle\frac{x}{\sqrt{x^2+1}}=\displaystyle\frac{x\cos(\sqrt{x^2+1})}{\sqrt{x^2+1}}$