Question

How do you differentiate `tan(sin x)`?

Answer 1

`sec^2(sinx)cosx`

Explanation:

Given: `d/dx(tan(sinx))`.

Here, let `y=tan(sinx)`.

Use the chain rule, which states that,

`dy/dx=dy/(du)*(du)/dx`

Let `u=sinx,:.(du)/dx=cosx`.

Then `y=tanu,dy/(du)=sec^2u`.

So, combining our results, we get:

`dy/dx=sec^2u*cosx`

Substituting back `u=sinx`, we get:

`dy/dx=sec^2(sinx)*cosx`

`=sec^2(sinx)cosx`