# 18.  Find the derivative of the following functions (it is to be understood that a, b, c, d, p, q, r and s are fixed non-zero constants and m and n are integers):         $\frac{\sec x -1}{\sec x +1}$

P Pankaj Sanodiya

Given,

$f(x)=\frac{\sec x -1}{\sec x +1}$

which also can be written as

$f(x)=\frac{1-\cos x}{1+\cos x}$

Now,

As we know the derivative of such function

$\frac{d(\frac{y_1}{y_2})}{dx}=\frac{y_2d(\frac{dy_1}{dx})-y_1(\frac{dy_2}{dx})}{y_2^2}$

So, The derivative of the function is,

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{(1+\cos x)d(\frac{d(1-\cos x)}{dx})-(1-\cos x)(\frac{d(1+\cos x)}{dx})}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{(1+\cos x)(-(-\sin x))-(1-\cos x)(-\sin x)}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{\sin x+\sin x\cos x+\sin x-\cos x\sin x}{(1+\cos x)^2}$

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{2\sin x}{(1+\cos x)^2}$

Which can also be written as

$\frac{d(\frac{1-\cos x}{1+\cos x})}{dx}=\frac{2\sec x\tan x}{(1+\sec x)^2}$.

Exams
Articles
Questions