# How do you find the derivative of sin^2(x)?

May 24, 2016

${y}^{'} = 2 \cdot \sin x \cdot \cos x$

#### Explanation:

$y = a \cdot b$

$y ' = a ' \cdot b + b ' \cdot a$

$y = {\sin}^{2} = \sin x \cdot \sin x$

$\frac{d}{d x} \sin x = \cos x$

$y ' = \cos x \cdot \sin x + \cos x \cdot \sin x$

${y}^{'} = 2 \cdot \sin x \cdot \cos x$