Question

If $y = {\tan ^{ - 1}}\left( {\sec x + \tan x} \right)$, then $\dfrac{{dy}}{{dx}} =$
(A) $1$
(B) $\dfrac{1}{2}$
(C) $- 1$
(D) $0$

Answer 1

In this question, we have to choose the required solution for the given particular options. The given of the question is the value of the given term. We have to find the first derivative term of the given term. For that, first we have to differentiate the given term with respect to x. By using the chain rule we can differentiate the given particular term to find the solution, then we will get the required result.

Formula used: In calculus, the chain rule is a formula to compute the derivative of a composite function.
If f is a composite function, it can be constructed as $y = f\left( {g\left( x \right)} \right)$. In this case, the chain rule states that the derivative of a function $y = f\left( {g\left( x \right)} \right)$ is defined as: $\dfrac{{dy}}{{dx}} = f'\left( {g\left( x \right)} \right)g'\left( x \right)$
Differentiation formula:
$\dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$
$\dfrac{d}{{dx}}\left( {{{\tan }^{ - 1}}x} \right) = \dfrac{1}{{1 + {x^2}}}$
$\dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
Trigonometric formula:
${\sec ^2}\theta - {\tan ^2}\theta = 1$
Algebraic formula:
${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$

Complete step-by-step answer:
It is given that, $y = {\tan ^{ - 1}}\left( {\sec x + \tan x} \right)$.
We need to find out the value of $\dfrac{{dy}}{{dx}}$.
Now, $y = {\tan ^{ - 1}}\left( {\sec x + \tan x} \right)$.
Differentiating both sides with respect to x, we get,
$\dfrac{{dy}}{{dx}} = \dfrac{1}{{1 + {{\left( {\sec x + \tan x} \right)}^2}}} \times \left( {\sec x\tan x + {{\sec }^2}x} \right)$ [using chain rule]
Using the algebraic formula, ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$
$\Rightarrow \dfrac{{\left( {\sec x\tan x + {{\sec }^2}x} \right)}}{{1 + \left( {{{\sec }^2}x + {{\tan }^2}x + 2\sec x\tan x} \right)}}$
Applying the trigonometric formula, ${\sec ^2}\theta - {\tan ^2}\theta = 1$
$\Rightarrow \dfrac{{\left( {\sec x\tan x + {{\sec }^2}x} \right)}}{{{{\sec }^2}x - {{\tan }^2}x + {{\sec }^2}x + {{\tan }^2}x + 2\sec x\tan x}}$
Add and subtract the terms we get,
$\Rightarrow \dfrac{{\left( {\sec x\tan x + {{\sec }^2}x} \right)}}{{2{{\sec }^2}x + 2\sec x\tan x}}$
Taking the common term 2 outside from the brackets,
$\Rightarrow \dfrac{{\left( {\sec x\tan x + {{\sec }^2}x} \right)}}{{2\left( {\sec x\tan x + {{\sec }^2}x} \right)}}$
Cancelling the similar terms in numerator and denominator, we get,
$\Rightarrow \dfrac{1}{2}$.
Hence, we get, $\dfrac{{dy}}{{dx}} = \dfrac{1}{2}$

$\therefore$ The option (B) is the correct option.

Note: The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus.
The first Derivative of a function $y = f\left( x \right)$can be written as $\dfrac{{dy}}{{dx}}$ or $f'\left( x \right)$.
Composite function:
In mathematics, function composition is an operation that takes two functions f and g produces a function y such that $y(x) = f\left( {g\left( x \right)} \right)$.