Derivative of Tangent Functions

Problem

Figure 1. Trigonometric ratios visualized

Let the rational function $\frac{f(x)}{g(x)}$ be composed of two differentiable functions $f(x)$and $g(x)$, where $g(x) \ne 0$. The quotient rule states that the derivative of this rational function is $\frac{d}{dx} \left [ \frac{f(x)}{g(x)} \right ] = \frac{g(x)\frac{d}{dx}[f(x)] - f(x) \frac{d}{dx}[g(x)]}{[g(x)]^2}$. More compactly, the quotient rule can be expressed as $\left( \frac{f}{g} \right)' = \frac{gf'- fg'}{g^2}$.

Use the quotient rule to show that $\frac{d}{dx} \tan x = \sec^2 x$.

Hints

(1) $\tan x = \frac{\sin x}{\cos x}$

(2) $\cos^2 x +\sin^2 x = 1$

(3) $\sec x = \frac{1}{\cos x}$

Solution

Begin with the quotient identity of the tangent function (Hint 1), then apply the quotient rule of derivatives.

$$\frac{d}{dx} \tan x = \frac{d}{dx} \left[\frac{\sin x }{\cos x}\right]$$

$$\begin{align} \frac{d}{dx} \left[\frac{\sin x}{\cos x}\right] &= \frac{{\cos x} \cdot \frac{d}{dx}\sin x-\sin x \cdot \frac{d}{dx}\cos x}{\cos^2 x} \\[5 pt] &= \frac{\cos x \cdot \cos x -\sin x \cdot (-\sin x)}{\cos^2 x} \\[5 pt] &= \frac{\cos^2 x +\sin^2 x }{\cos^2 x}. \end{align}$$

Since $\cos^2 x +\sin^2 x = 1$, we get

$$\frac{d}{dx} \left[\frac{\sin x}{\cos x}\right] = \frac{1}{\cos^2 x}$$

and because $\sec x = \frac{1}{\cos x}$, we get

$$\frac{d}{dx} \tan x = \sec^2 x.$$