Derivative of tan x by Product, Quotient and Chain Rule

The derivative of $\tan x$ is $\sec^2 x$, that is, $\dfrac{d}{dx}(\tan x)=\sec^2 x$. In this post, we will learn how to find the derivative of \tan x by the product rule of derivatives as well as the quotient and chain rule of derivatives. At first, we will recall the rules.

Let $f(x)$ and $g(x)$ be two functions of $x$. Then the derivative of the product $fg$ and of the quotient $f/g$with respect to $x$ are given as follows:

Product and Quotient Rule of Derivatives

$\dfrac{d}{dx}(fg)=f \dfrac{dg}{dx} + g \dfrac{df}{dx}$ $\cdots (i)$
$\dfrac{d}{dx}(f/g)=\dfrac{g\dfrac{df}{dx}-f\dfrac{dg}{dx}}{g^2}$ $\cdots (ii)$

Derivative of tanx by Product Rule

We know that $\tan x=\sin x/ \cos x$

As $\dfrac{1}{\cos x} =\sec x,$ we get that

$\tan x =\sin x \sec x$

Differentiating $\tan x$ with respect to $x$ by the above product rule (i) of derivatives, we have

$\dfrac{d}{dx}(\tan x)$ $=\sin x \dfrac{d}{dx}(\sec x)+$ $\sec x \dfrac{d}{dx}(\sin x)$

$=\sin x \sec x \tan x + \sec x \cos x$

$=\sin x \cdot \dfrac{1}{\cos x} \cdot \dfrac{\sin x}{\cos x}+$ $\dfrac{1}{\cos x} \cdot \cos x$

$=\dfrac{\sin^2 x}{\cos^2 x}+1$

$=\dfrac{\sin^2 x+ \cos^2 x}{\cos^2 x}$

$=\dfrac{1}{\cos^2 x}$ [as $\sin^2 x+\cos ^2 x=1$]

$=\sec^2 x$

So the derivative of $\tan x$ is $\sec^2 x.$

Derivative of tan x by Quotient Rule

We have $\tan x=\sin x/ \cos x$

Differentiating $\tan x$ with respect to $x$ by the above quotient rule (ii) of derivatives, we obtain that

$\dfrac{d}{dx}(\tan x)=\dfrac{d}{dx}(\dfrac{\sin x}{\cos x})$

$=\dfrac{\cos x \dfrac{d}{dx}(\sin x)-\sin x \dfrac{d}{dx}(\cos x)}{\cos^2 x}$

$=\dfrac{\cos x \cos x-\sin x (-\sin x)}{\cos^2 x}$

$=\dfrac{\cos^2 x +\sin^2 x}{\cos^2 x}$

$=\dfrac{1}{\cos^2 x}$ $=\sec^2 x.$

Derivative of tan x by Chain Rule

Let $z = \cot x$ .

We have

$\dfrac{d}{dx}(\tan x) = \dfrac{d}{dx}(\dfrac{1}{\cot x})$

$\Rightarrow \dfrac{d}{dx}(\tan x) = \dfrac{d}{dx}(\dfrac{1}{z})$

$\Rightarrow \dfrac{d}{dx}(\tan x) = \dfrac{d}{dz}(\dfrac{1}{z}) \cdot \dfrac{dz}{dx}$ by the chain rule of derivatives

$\Rightarrow \dfrac{d}{dx}(\tan x) = -\dfrac{1}{z}^2$ $\cdot (-\text{cosec}^2 x)$  as $z=\cot x$

$\Rightarrow \dfrac{d}{dx}(\tan x)=\dfrac{1}{\cot^2 x}$ $\cdot \text{cosec}^2 x$

$\Rightarrow \dfrac{d}{dx}(\tan x) = \dfrac{\sin^2 x}{\cos^2 x} \cdot \dfrac{1}{\sin^2 x}$

$=\dfrac{1}{\cos^2 x}$

$=\sec^2 x.$

Thus the derivative of $\tan x$ is $\sec^2 x$ obtained by the chain rule of derivatives.