Question

How do you differentiate `f(x)=(cos x+ 4sin^2x^2)^6` using the chain rule?

Answer 1

`6(cosx+4sin^2x^2)^5(-sinx+16xsinx^2cosx^2)`

Explanation:

In several stages.

First stage - chain rule for the bracket as a whole:
`d/dx[(cosx+4sin^2x^2)^6]`
`=`
`6(cosx+4sin^2x^2)^5*d/dx[cosx+4sin^2x^2]`

Second stage - two terms, first one simple, second chain rule again:
`d/dx[cosx]=-sinx`
`d/dx[4sin^2x^2]=8sinx^2 *d/dx [sinx^2]`

Third stage - chain rule again
`d/dx[sinx^2]=cosx^2*d/dx[x^2]`

Fourth stage - simple differentiation
`d/dx[x^2]=2x`

Work back through the process:
`d/dx[x^2]=2x`
`d/dx[sinx^2]=2xcosx^2`
`d/dx[4sin^2x^2]=16xsinx^2cosx^2`

`d/dx[(cosx+4sin^2x^2)^6]`
`=`
`6(cosx+4sin^2x^2)^5(-sinx+16xsinx^2cosx^2)`