What is the derivative of tan(x)?

The Mathematician — Thu, 08 Sep 2022 13:00:12 +0000

We will find out what the derivative of

\tan(x)

is. We will be doing that by the fact that

\tan(x) = \sin(x)/\cos(x)

and apply the quotient rule. We saw earlier a difficult version how to show the derivative of

\tan(x)

with first principle method, but in this article, we take it more easier.

Proof. Let

h(x) = \tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{f(x)}{g(x)}

. Then

\begin{align*} h'(x) = \frac{d}{dx} \bigg(\frac{f(x)}{g(x)}\bigg) = \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}. \end{align*}

We have seen that here that

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

and here that

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

. Therefore, we get

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

Substituting everything together, we get

\begin{align*} h'(x) &= \frac{d}{dx} \bigg(\frac{f(x)}{g(x)}\bigg) \\ &= \frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2} \\ &= \frac{\cos(x)\cos(x)-\sin(x)(-\sin(x))}{\cos^2(x)} \\ &= \frac{\cos^2(x) + \sin^2(x)}{\cos^2(x)} \\ &= \frac{1}{\cos^2(x)} \\ &= \sec^2(x) \end{align*}

Therefore, we get

h'(x) = \frac{d}{dx} \tan(x) = \frac{1}{\cos^2(x)} = \sec^2(x)

derivative of tan(x) Archives - Epsilonify