What is the derivative of sin(2x)?

The Mathematician — Fri, 09 Sep 2022 13:00:16 +0000

We will find out the derivative of

\sin(2x)

.

Proof. Let

F(x) = \sin(2x)

f(u) = \sin(u)

and

g(x) = 2x

such that

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

. We are going to use the chain rule here:

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

We have seen here that

\frac{d}{du}\sin(u) = \cos(u)

, and

\frac{d}{dx} 2x = 2

. Therefore, we get

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

Plugging everything together, we get

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

So the derivative of

\sin(2x)

2\cos(2x)

What is the derivative of sin^2(x)

The Mathematician — Tue, 06 Sep 2022 15:00:31 +0000

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)