
Find the differentiation of the following function ${\sec ^{ - 1}}\tan x.$
Answer
596.4k+ views
Hint- In order to differentiate the given term, we have to use chain rule for differentiation of given expression, we will use the formula of differentiation given below
$\dfrac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{x\sqrt {{x^2} - 1} }}$ then simplify it, we will get the answer.
Complete step-by-step answer:
Given term ${\sec ^{ - 1}}\tan x.$
Let us consider $f\left( x \right) = {\sec ^{ - 1}}\tan x$
As we know that the differentiation of \[{\sec ^{ - 1}}x\] is given as
$\dfrac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{x\sqrt {{x^2} - 1} }}$
In order to solve the problem we will use the chain rule.
According to the chain rule, we have:
$\dfrac{{dp}}{{dq}} = \dfrac{{dp}}{{du}}.\dfrac{{du}}{{dq}}$
Using the chain rule let us proceed
So, by using the formula, we get
$
\dfrac{{df\left( x \right)}}{{dx}} = \dfrac{{d\left( {{{\sec }^{ - 1}}\tan x} \right)}}{{dx}} \\
= \dfrac{1}{{\tan x\sqrt {{{\tan }^2}x - 1} }} \times \dfrac{{d\tan x}}{{dx}} \\
\dfrac{1}{{\tan x\sqrt {{{\tan }^2}x - 1} }} \times {\sec ^2}x{\text{ }}\left[ {\because \dfrac{{d\tan x}}{{dx}} = {{\sec }^2}x} \right] \\
= \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}\sqrt {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1} }} \times \dfrac{1}{{{{\cos }^2}x}} \\
= \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}\sqrt {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}}} }} \times \dfrac{1}{{{{\cos }^2}x}} \\
$
By simplifying above equation, we get
\[
= \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}\dfrac{{\sqrt {{{\sin }^2}x - {{\cos }^2}x} }}{{\cos x}}}} \times \dfrac{1}{{{{\cos }^2}x}} \\
= \dfrac{1}{{\sin x\sqrt {{{\sin }^2}x - {{\cos }^2}x} }} \\
\]
Hence the differentiation of ${\sec ^{ - 1}}\tan x$ is \[\dfrac{1}{{\sin x\sqrt {{{\sin }^2}x - {{\cos }^2}x} }}\]
Note- In order to solve these types of questions, first of all remember all the properties of differentiation and learn about quotient rule and product rule of differentiation. You must also be aware of the chain rule of the differentiation. You must also have good knowledge of topics like limits and continuity. Keep in mind that continuous functions are differentiable.
$\dfrac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{x\sqrt {{x^2} - 1} }}$ then simplify it, we will get the answer.
Complete step-by-step answer:
Given term ${\sec ^{ - 1}}\tan x.$
Let us consider $f\left( x \right) = {\sec ^{ - 1}}\tan x$
As we know that the differentiation of \[{\sec ^{ - 1}}x\] is given as
$\dfrac{{d\left( {{{\sec }^{ - 1}}x} \right)}}{{dx}} = \dfrac{1}{{x\sqrt {{x^2} - 1} }}$
In order to solve the problem we will use the chain rule.
According to the chain rule, we have:
$\dfrac{{dp}}{{dq}} = \dfrac{{dp}}{{du}}.\dfrac{{du}}{{dq}}$
Using the chain rule let us proceed
So, by using the formula, we get
$
\dfrac{{df\left( x \right)}}{{dx}} = \dfrac{{d\left( {{{\sec }^{ - 1}}\tan x} \right)}}{{dx}} \\
= \dfrac{1}{{\tan x\sqrt {{{\tan }^2}x - 1} }} \times \dfrac{{d\tan x}}{{dx}} \\
\dfrac{1}{{\tan x\sqrt {{{\tan }^2}x - 1} }} \times {\sec ^2}x{\text{ }}\left[ {\because \dfrac{{d\tan x}}{{dx}} = {{\sec }^2}x} \right] \\
= \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}\sqrt {\dfrac{{{{\sin }^2}x}}{{{{\cos }^2}x}} - 1} }} \times \dfrac{1}{{{{\cos }^2}x}} \\
= \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}\sqrt {\dfrac{{{{\sin }^2}x - {{\cos }^2}x}}{{{{\cos }^2}x}}} }} \times \dfrac{1}{{{{\cos }^2}x}} \\
$
By simplifying above equation, we get
\[
= \dfrac{1}{{\dfrac{{\sin x}}{{\cos x}}\dfrac{{\sqrt {{{\sin }^2}x - {{\cos }^2}x} }}{{\cos x}}}} \times \dfrac{1}{{{{\cos }^2}x}} \\
= \dfrac{1}{{\sin x\sqrt {{{\sin }^2}x - {{\cos }^2}x} }} \\
\]
Hence the differentiation of ${\sec ^{ - 1}}\tan x$ is \[\dfrac{1}{{\sin x\sqrt {{{\sin }^2}x - {{\cos }^2}x} }}\]
Note- In order to solve these types of questions, first of all remember all the properties of differentiation and learn about quotient rule and product rule of differentiation. You must also be aware of the chain rule of the differentiation. You must also have good knowledge of topics like limits and continuity. Keep in mind that continuous functions are differentiable.
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