Derivatives and Trig Identities

Let's find the first derivative of $\displaystyle \sin^2(x)$ . We use the chain rule, that is that the derivative of $\displaystyle u(v(x)) = u'(v(x))v'(x)$ . Let $\displaystyle u(x) = x^2$ and $\displaystyle v(x) = \sin(x)$ .

$\displaystyle u'(v(x)) = 2\sin(x)$ and $\displaystyle v'(x) = \cos(x)$ .

$\displaystyle u'(v(x))v'(x) = 2\sin(x)\cos(x)$ .

Now we take the derivative of $\displaystyle 2\sin(x)\cos(x)$ , using the product rule, or that the derivative of $\displaystyle u(x)v(x) = u(x)v'(x) + v(x)u'(x)$ .

$\displaystyle u(x) = 2\sin(x)$ and $\displaystyle v(x) = \cos(x)$

$\displaystyle u(x)v'(x) + v(x)u'(x) = -2\sin(x)\sin(x) + 2\cos(x)\cos(x) = 2(\cos^2(x)-\sin^2(x)) = 2\cos(2x)$

The answer is $\displaystyle 2\cos(2x)$