Question

Find the derivative of y=sin^2x
The answer is sin2x, but I don't get why that is.

Answer 1

ok... so let

u = sin(x) du/dx = cos(x)

so

y = u^2 then dy/du = 2u

then using the chain rule

\[\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\]

which is

\[\frac{dy}{dx} = 2u \times \cos(x)\]

make the substitution u = sin(x) so its

\[\frac{dy}{dx} = 2\sin(x) \times \cos(x) = 2\sin(x)\cos(x) = \sin(2x)\]

hope it helps

Answer 2

\[
\frac{d}{dx}\sin^2(x) = \frac{d}{dx}(\sin(x))^2 =
2(\sin(x))^{2-1} * \frac{d}{dx}(\sin(x) =
2\sin(x)*\cos(x) = \sin(2x)
\]