Question

How do you use the chain rule to differentiate `(-cosx)^2008`?

Answer 1

See below.

Explanation:

The chain rule states that:

`dy/dx=(dy)/(du)*(du)/dx` where `u =f(x)`

Let `u = -cosx`

Then we have:

`y=u^2008`

`dy/(du)(u^(2008))=2008u^(2007)`

`(du)/dx(-cosx)=sinx`

`:.`

`dy/dy=(dy)/(du)*(du)/dx=2008u^(2007)*sinx`

But `u = -cosx`

So:

`2008(-cosx)^(2007)*sinx=-2008sinx(cosx)^(2007)`