Question

What is the derivative of ${\sec ^{ - 1}}(x)$ ?

Answer 1

Hint : Here, the given question has a trigonometric function. We have to find the derivative or differentiated term of the function. First consider the function $y$, then differentiate $y$ with respect to $x$ by using a standard differentiation formula of trigonometric ratio and use chain rule for differentiation. And on further simplification we get the required differentiate value.
Formula used:
In the trigonometry we have standard differentiation formula
the differentiation of cos x is -sin x that is $\dfrac{d}{{dx}}(\sec x) = \sec x.\tan x$

Complete step by step solution:
The differentiation of a function is defined as the derivative or rate of change of a function. The function is said to be differentiable if the limit exists.
Consider the given function
$\Rightarrow y = {\sec ^{ - 1}}(x)$ ---------- (1)
Take $\sec$ on both sides we have
$\Rightarrow \sec y = \sec \left( {{{\sec }^{ - 1}}(x)} \right)$
On simplifying we get
$\Rightarrow \sec y = x$ ----------(2)
Differentiate function y with respect to x
$\Rightarrow \dfrac{d}{{dx}}\left( {\sec y} \right) = \dfrac{d}{{dx}}\left( x \right)$ -------(3)
As we know the formula $\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$, then
Equation (2) becomes
$\Rightarrow \sec y.\tan y\dfrac{{dy}}{{dx}} = 1$ ---------- (4)
As we know the trigonometric identities which is given as $1 + {\tan ^2}y = {\sec ^2}y$, by the equation (2) the trigonometric identity
$\Rightarrow 1 + {\tan ^2}y = {x^2}$
Take 1 to the RHS
$\Rightarrow {\tan ^2}y = {x^2} - 1$
Take square root on both sides and it is written as
$\Rightarrow \tan y = \sqrt {{x^2} - 1}$ -----------(5)
By using the equation (5) and the equation (2), the equation (4) is written as
$\Rightarrow x\sqrt {{x^2} - 1} \dfrac{{dy}}{{dx}} = 1$
Take $x\sqrt {{x^2} - 1}$ to the RHS and it is written as
$\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{x\sqrt {{x^2} - 1} }}$
Hence, it’s a required differentiated value.
So, the correct answer is “ $\dfrac{1}{{x\sqrt {{x^2} - 1} }}$ ”.

Note : The student must know about the differentiation formulas for the trigonometry ratios and these differentiation formulas are standard. If the function is a product of two terms and the both terms are the function of x then we use the product rule of differentiation to the function.