Question

Question #321de

Answer 1

`e^sinx(cos^2x)-e^sinx(sinx)`

Explanation:

We know that the first derivative is `e^sinx(cosx)`

To get the second derivative we apply the product rule:

Where `f'(g)+g'(f)`

Let's say that `f` is our `e^sinx` and `g` is `cosx`

We take the derivative of `f` and multiply it by `g` and add the derivative of `g` multiplied by `f`

`e^sinx(cosx)(cosx)+(-sinx)(e^sinx)`

Simplify:

`e^sinx(cos^2x)-e^sinx(sinx)`