# Differentiate

Question:

Differentiate

$\left(\frac{\sec x-\tan x}{\sec x+\tan x}\right)$

Solution:

To find: Differentiation of $\left(\frac{\sec x-\tan x}{\sec x+\tan x}\right)$

Formula used: (i) $\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

(ii) $\frac{d \sec x}{d x}=\sec x \tan x$

(iii) $\frac{d \tan x}{d x}=\sec ^{2} x$

Let us take $u=(\sec x-\tan x)$ and $v=(\sec x+\tan x)$

$u^{\prime}=\frac{d u}{d x}=\frac{d(\sec x-\tan x)}{d x}=\left(\sec x \tan x-\sec ^{2} x\right)$

$v^{\prime}=\frac{d v}{d x}=\frac{d(\sec x+\tan x)}{d x}=\left(\sec x \tan x+\sec ^{2} x\right)$

Putting the above obtained values in the formula:-

$\left(\frac{u}{v}\right)^{\prime}=\frac{u^{\prime} v-u v^{\prime}}{v^{2}}$ where $v \neq 0$ (Quotient rule)

$\left[\frac{\sec x-\tan x}{\sec x+\tan x}\right]=\frac{\left(\sec x \tan x-\sec ^{2} x\right)(\sec x+\tan x)-(\sec x-\tan x)\left(\sec x \tan x+\sec ^{2} x\right)}{(\sec x+\tan x)^{2}}$

$=\frac{\left(\sec x \tan x-\sec ^{2} x\right)(\sec x+\tan x)-(\sec x-\tan x)(\sec x)(\tan x+\sec x)}{(\sec x+\tan x)^{2}}$

$=\frac{(\sec x+\tan x)\left[\left(\sec x \tan x-\sec ^{2} x\right)-(\sec x-\tan x)(\sec x)\right]}{(\sec x+\tan x)^{2}}$

$=\frac{(\sec x+\tan x)\left[\left(\sec x \tan x-\sec ^{2} x\right)-\left(\sec ^{2} x-\sec x \tan x\right)\right]}{(\sec x+\tan x)^{2}}$

$=\frac{(\sec x+\tan x)\left[2 \sec x \tan x-2 \sec ^{2} x\right]}{(\sec x+\tan x)^{2}}$

$=\frac{2 \sec x[\tan x-\sec x]}{(\sec x+\tan x)}$

Ans) $=\frac{2 \sec x[\tan x-\sec x]}{(\sec x+\tan x)}$