Question

How do you differentiate `f(x)=tan(e^(x-x^2))` using the chain rule?

Answer 1

`f'(x)=e^(x-x^2)*(1-2x)*sec^2(e^(x-x^2))`

Explanation:

Note that

`(tan(x))'=sec^2(x)`
so we get by the chain rule

`sec^2(e^(x-x^2))*e^(x-x^2)*(1-2x)`