Differentiate: f(x)=2(sin(2x))^2 with respect to x, and evaluate as a single trigonometric function.

f(x) = 2sin2(2x)Therefore, using the chain rule: f'(x)=2 x 2cos(2x) x 2sin(2x)(The 2 at the front arises from the constant 2, at the start of f(x), the 2cos(2x) comes from differentiating sin2(2x), then the 2sin(2x) comes from decreasing the original power of the sine function by 1 and multiplying by the constant in the function, 2)Therefore, f'(x)=6cos(2x)sin(2x)As we know 2sin(x)cos(x)=sin(2x) (double-angle formula), we can simplify f'(x) into f'(x)=3sin(4x)

SH

Related Maths A Level answers

All answers ▸

Differentiate y=x^3*(x^2+1)


Find all solutions of the equation in the interval [0, 2π]. 5 cos^3 x = 5 cos x


Find the gradient of y=6x^3+2x^2 at (1,1)


Differentiate with respect to x: F(x)=(x^2+1)^2