Question

How do you find the derivative of `y= sin{cos^2(tanx)}`?

Answer 1

`(dy)/(dx)=-2sec^2(x)sin(tanx)cos(tanx)cos(cos^2(tan(x)))`

Explanation:

`y= sin(cos^2(tanx))=sin(f(x))`

`y=sin(f(x))=>(dy)/(dx)=f'(x)cos(f(x))`

`f(x)=cos^2(g(x))=(cos(g(x)))^2`
`f'(x)=2*d/(dx)[cos(g(x))] * cos(g(x))`
`f'(x)=2*-g'(x)sin(g(x))*cos(g(x))`
`f'(x)=-2g'(x)sin(g(x))cos(g(x))`

`g(x)=tanx`
`g'(x)=sec^2x`

Therefore, `f'(x)=-2sec^2(x)sin(tanx)cos(tanx)`

`(dy)/(dx)=-2sec^2(x)sin(tanx)cos(tanx)cos(cos^2(tan(x)))`