Question

How do you differentiate `f(x)=sin(e^(3-x))` using the chain rule?

Answer 1

`f'(x)=-e^(3-x)cos(e^(3-x))`

Explanation:

The Chain Rule, when applied to the sine, tells us that

`d/dxsin(u)=cosu*(du)/dx,` where `u` is some function in terms of `x.`

Here, `u=e^(3-x)`, so

`f'(x)=cos(e^(3-x))*d/dxe^(3-x)`

`d/dxe^(3-x)=e^(3-x)*d/dx(3-x)=-e^(3-x)` so we get

`f'(x)=-e^(3-x)cos(e^(3-x))`