Question

What is the derivative of `sin(2x)cos(2x)`?

Answer 1

`2\cos(4x)`

Explanation:

Given function:

`\sin (2x)\cos (2x)`

`1/2(2\sin (2x)\cos (2x))`

`1/2\sin (4x)`

Differentiating given function w.r.t. `x` as follows

`d/dx(1/2\sin(4x))`

`=1/2\d/dx(\sin(4x))`

`=1/2\cos(4x)d/dx(4x)`

`=1/2\cos(4x)(4)`

`=2\cos(4x)`

Answer 2

`2cos4x`

Explanation:

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

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

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

`f(x)=sin2xrArrf'(x)=2cos2x`

`g(x)=cos2xrArrg'(x)=-2sin2x`

`d/dx(sin2xcos2x)`

`=-2sin^2 2x+2cos^2 2x`

`=2(cos^2 2x-sin^2 2x)=2cos4x`