What is the Derivative of sin^2(x)?

We will determine the derivative of

\sin^2(x)

. We will do that in two different ways: chain rule and product rule

Proof 1. Let

F(x) = \sin^2(x)

f(u) = u^2

and

g(x) = \sin(x)

such that

F(x) = f(g(x))

. We will use the chain rule:

\begin{align*} F'(x) = f'(g(x))g'(x). \end{align*}

We do know from this article that

(\sin(x))' = \cos(x)

. So

\begin{align*} f'(g(x)) = 2g(x) = 2\sin(x) \quad \text{and} \quad g'(x) = \cos(x). \end{align*}

So we get

\begin{align*} F'(x) = f'(g(x))g'(x) = 2\sin(x)\cos(x). \end{align*}

Proof 2. Let

f(x) = \sin^2(x)

. Then we want to show

\begin{align*} f'(x) = 2\cos(x)\sin(x). \end{align*}

The first step we apply is that

\begin{align*} \sin^2(x) = \sin(x)\sin(x). \end{align*}

This must ring a bell because we can apply the product rule. So we get

\begin{align*} (\sin(x)\sin(x))' = (\sin(x))'\sin(x) + \sin(x)(\sin(x))'. \end{align*}

As we know from this article that

(\sin(x))' = \cos(x)

, we get

\begin{align*} (\sin(x))'\sin(x) + \sin(x)(\sin(x))' = \cos(x)\sin(x) + \sin(x)\cos(x). \end{align*}

This implies that

\cos(x)\sin(x) + \sin(x)\cos(x)\ = 2\cos(x)\sin(x)

. In a complete picture, we get

\begin{align*} h'(x) &= (\sin(x)\sin(x))' \\ &= (\sin(x))'\sin(x) + \sin(x)(\sin(x))' \\ &= \cos(x)\sin(x) + \sin(x)\cos(x) \\ &= 2\cos(x)\sin(x). \end{align*}

Therefore,

f'(x) = \frac{d}{dx} \sin^2(x) = 2\cos(x)\sin(x)

What is the derivative of sin^2(x)