Differentiate 2sin^2xcos^2x

Answer 1

Recall that sin(2θ) = 2sinθcosθ

So, 2sin²xcos²x = 2(sinxcosx)² = 2[(1/2)2sinxcosx]² = (1/2)sin²(2x)

By the Chain Rule, the derivative is (1/2)[2sin(2x)](sin(2x)}' = sin(2x)cos(2x)(2) = 2sin(2x)cos(2x) = sin(4x)

Answer 2

Given that f(x) = 2sin²(x)cos²(x) we can do it one of two ways...

First way: Use Trig Identity

2sin²(x)cos²(x) = 2sin(x)cos(x)*sin(x)cos(x) = sin(2x)*(1/2)sin(2x) = (1/2)*sin²(2x)

f(x) = (1/2)*sin²(2x)

f'(x) = 2*(1/2)*sin(2x) * (2)cos(2x) = 2sin(2x)cos(2x) = sin(4x)

Second way: Brute force

f(x) = 2sin²(x)cos²(x) use product rule

g(x) = sin²(x)

h(x) = cos²(x)

f'(x) = g(x)*h'(x) + g'(x)*h(x)

g'(x)= 2sin(x)cos(x)

h'(x) = -2sin(x)cos(x)

f'(x) = 2 * [sin²(x) * -2sin(x)cos(x) + 2sin(x)cos(x) * cos²(x) ]

f'(x) = 4sin(x)cos(x) * [ cos²(x) -sin²(x) ]

f'(x) = 2sin(2x) * cos(2x)

f'(x) = sin(4x)