Question

What is the derivative of `f(x)=cos(pi(x-1)^2/x^2)`?

Answer 1

`-(2pisin(pi-(2pi)/x+pi/x^2))/x^2+(2pisin(pi-(2pi)/x+pi/x^2))/x^3`

Explanation:

Put this into a more friendly form by rearranging. Expand out the inside of the brackets and leave things in coefficient-`x`-power form to make differentiating much easier.

`cos(pi(x-1)^2/x^2)=cos(pi(x^2-2x+1)/x^2)`

`=cos(pi(1-2x^-1+x^-2))`

Now you should use the chain rule, which says that if

`f(x)=g(h(x))` (one function embedded inside another), then

`f'(x)=h'(x)*g'(h)`

If we say that `h(x)=pi(1-2x^-1+x^-2)` and `g(h)=cos(h)`, then

`h'(x)=pi(2x^-2-2x^-3)`, and

`g'(h)=-sin(h)=-sin(pi(1-2x^-1+x^-2))`

Putting all of this together as it says in the chain rule,

`f'(x)=pi(2x^-2-2x^-3)*-sin(pi(1-2x^-1+x^-2))`

We can expand this out and rearrange in a nicer form using laws of indices to get

`f'(x)=-((2pi)/x^2-(2pi)/x^3)sin(pi-(2pi)/x+pi/x^2)`

`=-(2pisin(pi-(2pi)/x+pi/x^2))/x^2+(2pisin(pi-(2pi)/x+pi/x^2))/x^3`