Question

How do you find the derivative of `h(x)= sin2xcos2x`?

Answer 1

We must use the product rule here, which states that, be `y=f(x)g(x)`, then `y'=f'(x)g(x)+f(x)g'(x)`

Thus, let's just bear in mind the derivatives of these trigonometric functions:

• `y=sin(u)`, then `y'=u'cosu`
• `y=cos(u)`, then `y'=-u'sinu`

`(dy)/(dx)=2cos2x(cos2x)+(sin2x)(-2sin2x)`

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