What is the derivative of sec(x)?

The Mathematician — Fri, 16 Sep 2022 13:00:37 +0000

We will use the chain rule to show that the derivative of

\sec(x)

\tan(x)\cos(x)

.

Proof. Let

F(x) = \sec(x) = \frac{1}{\cos(x)}

f(u) = \frac{1}{u}

and

g(x) = \cos(x)

such that

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

. Then we will use the chain rule:

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

We know from this article that

g'(x) = \frac{d}{dx} \cos(x) = -\sin(x)

and

f'(u) = \frac{d}{du} \frac{1}{u} = \frac{-1}{u^2}

. So we get

\begin{align*} f'(g(x)) = \frac{-1}{g(x)^2} = \frac{-1}{\cos^2(x)} \quad \text{and} \quad g'(x) = -\sin(x). \end{align*}

Substituting everything together, we get

\begin{align*} F'(x) &= f'(g(x))g'(x) \\ &= \frac{-1}{\cos^2(x)}\cdot(-sin(x)) \\ &= \frac{\sin(x)}{\cos^2(x)} \\ &= \frac{\sin(x)}{\cos(x)}\frac{1}{\cos(x)} \\ &= \tan(x) \frac{1}{\cos(x)} \\ &= \tan(x)\sec(x) \end{align*}

So we have that the derivative of

\sec(x)

\tan(x)\sec(x)

derivative of sec(x) Archives - Epsilonify