Question

What is the derivative of `(sinx)^2*(cosx)^2`?

Answer 1

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

Explanation:

You have to use the product rule to differentiate this expression.

`y=(sin x)^2xx(cos x)^2`

`u=(sin x)^2, v= (cos x)^2`

`u' = 2(sinx)^1cosx=2sinxcosx`

`v'=2(cosx)-sinx=-2cosxsinx`

`y'=vu'+uv'`

`y'=2sinxcos^3x-2cosxsin^3x=2sinxcosx(cos^2x-sin^2x)`

Answer 2

`d/dx sin^2xcos^2x = 1/2sin4x`

Explanation:

Another approach is to simply the expression using the identity:

`sin 2theta = 2sinthetacostheta`

before starting, which yields a simpler, (but identical) form of the solution:

`d/dx sin^2xcos^2x = d/dx (sinxcosx)^2`
`" "= d/dx (1/2sin2x)^2`
`" "= 1/4d/dx sin^2 2x`
`" "= 1/4*2sin2x*cos2x*2 \ \ \ ` (by the chain rule)
`" "= 1/2sin4x`