Question

What is the derivative of $\dfrac{1}{\sec x-\tan x}$?

Answer 1

We solve this problem by rationalising the denominator by using the standard identity of the trigonometric ratios of secant and tangent.
The identity of trigonometric ratios is given as,
${{\sec }^{2}}x-{{\tan }^{2}}x=1$
By using this identity we convert the given function completely to numerator and then we differentiate it to get the required answer. We use the formulas of derivatives given as,
$\dfrac{d}{dx}\left( \sec x \right)=\sec x.\tan x$
$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$

Complete step by step solution:
We are asked to find the derivative of the function $\dfrac{1}{\sec x-\tan x}$
Let us assume that the given function as,
$\Rightarrow f\left( x \right)=\dfrac{1}{\sec x-\tan x}$
Now, let us take the standard identity of the secant and tangent ratios that is,
${{\sec }^{2}}x-{{\tan }^{2}}x=1$
We know that the standard formula of algebra that is,
${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$

By using this formula to the standard identity then we get,
$\begin{aligned} & \Rightarrow \left( \sec x+\tan x \right)\left( \sec x-\tan x \right)=1 \\\ & \Rightarrow \dfrac{1}{\sec x-\tan x}=\sec x+\tan x \\\ \end{aligned}$
By using this in the given function then we get,
$\Rightarrow f\left( x \right)=\sec x+\tan x$
Now, let us differentiate the above equation on both sides with respect to $'x'$ then we get,
$\begin{aligned} & \Rightarrow \dfrac{d}{dx}\left( f\left( x \right) \right)=\dfrac{d}{dx}\left( \sec x+\tan x \right) \\\ & \Rightarrow {f}'\left( x \right)=\dfrac{d}{dx}\left( \sec x \right)+\dfrac{d}{dx}\left( \tan x \right) \\\ \end{aligned}$
We know that the standard formulas of derivatives of trigonometric ratios is given as,
$\dfrac{d}{dx}\left( \sec x \right)=\sec x.\tan x$
$\dfrac{d}{dx}\left( \tan x \right)={{\sec }^{2}}x$
By using these formulas for the above equation then we get,
$\begin{aligned} & \Rightarrow {f}'\left( x \right)=\sec x.\tan x+{{\sec }^{2}}x \\\ & \Rightarrow {f}'\left( x \right)=\sec x\left( \sec x+\tan x \right) \\\ \end{aligned}$
Therefore we can conclude that the derivative of the given function $\dfrac{1}{\sec x-\tan x}$ is given as,
$\therefore \dfrac{d}{dx}\left( \dfrac{1}{\sec x-\tan x} \right)=\sec x\left( \sec x+\tan x \right)$

Note: We need to note that when there is a possibility of rationalising the denominator of the given function then we need to do that first in order to make the calculations easy. Here, we are given that the function as $\dfrac{1}{\sec x-\tan x}$.
We know that we can directly differentiate the above function to get the answer. But that will take more time than usual. Here, as the denominator can be rationalised and can be converted into numerator only then by doing so that will reduce our time taking as well as calculations.