What is the derivative of cos^2(x)

The Mathematician — Wed, 07 Sep 2022 15:00:26 +0000

We will determine the derivative of

\cos^2(x)

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

Proof 1. Let

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

f(u) = u^2

and

g(x) = \cos(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 know from this article that

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

. Therefore, we get

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

So we get

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

Proof 2. Let

f(x) = \cos^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*} \cos^2(x) = \cos(x)\cos(x) \end{align*}

We can apply the product rule now and, therefore

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

We have seen earlier that this article that

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

. Therefore, we get

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

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

. In a complete picture, we get

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

Therefore,

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

derivative of cos x square Archives - Epsilonify