Question

What is the derivative of `tan(sin x)`?

Answer 1

`sec^2(sinx)cosx`

Explanation:

Use the chain rule.

According to the chain rule,

`d/dx(tan(u))=sec^(u)*(du)/dx`

Therefore,

`d/dx(tan(sinx))=sec^2(sinx)*d/dx(sinx)`

`=>sec^2(sinx)cosx`