Question

How do you find the derivative of `(cos^2(x)sin^2(x))`?

Answer 1

`1/2sin4x`

Explanation:

`"differentiate using the "color(blue)"product/chain rules"`

`"given "y=f(x)g(x)" then"`

`dy/dx=f(x)g'(x)+g(x)f'(x)larrcolor(blue)"product rule"`

`f(x)=cos^2xrArrf'(x)=2cosx(-sinx)`

`color(white)(xxxxxxxxxxxxxxx)=-2sinxcosx`

`g(x)=sin^2xrArrg'(x)=2sinx(cosx)`

`color(white)(xxxxxxxxxxxxxxx)=2sinxcosx`

`d/dx(cos^2xsin^2x)`

`=cos^2x(2sinxcosx)+sin^2x(-2sinxcosx)`

`=(2sinxcosx)(cos^2x-sin^2x)`

`=sin2xcos2x=1/2sin4x`