Question

Question #e54b4

Answer 1

`d/dx[5cos^2x]=-10cosxsinx`

Explanation:

The chain rule states that given the functions `y(u(x))` and `u(x)`, `d/dx[y]=d/(du)ytimesd/dxu`. Let's apply this to our problem.

First use the constant rule (`d/dx[au]=atimesd/dxu`) to simplify the problem.

`d/dx[5cos^2x]=5d/dx[cos^2x]`

Now it's time for the chain rule. Let `u=cosx` and let `y=u^2`.

`5d/dx[cos^2x]=5d/dx[u^2]=5d/dx[y]`

By the chain rule,

`d/dx[y]=d/(du)ytimesd/dxu=d/(du)u^2timesd/dxcosx`

Solving these derivatives,

`d/(du)u^2timesd/dxcosx=2utimes(-sinx)=-2usinx=-2cosxsinx`

So since `d/dx[y]=-2cosxsinx`, `5d/dx[y]=5(-2cosxsinx)=-10cosxsinx`
`d/dx[5cos^2x]=-10cosxsinx`