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.