Question:
If $y = \tan^{-1} ( \sec x - \tan x) $, then $ \frac{dy}{dx} = $
If $y = \tan^{-1} ( \sec x - \tan x) $, then $ \frac{dy}{dx} = $
Updated On: May 11, 2024
- $2$
- $-2$
- $ \frac{1}{2}$
- $ - \frac{1}{2}$
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The Correct Option is D
Solution and Explanation
$y = \tan^{-1} \left(\sec x-\tan x\right) $
$\Rightarrow \frac{dy}{dx} = \frac{1.\left(\sec x
\, \times \, \tan x \,-\, \sec^{2} x\right)}{1+\left(\sec x - \tan x\right)^{2}} $
$= \frac{\sec x \left( \tan x -\sec x \right)}{1+\sec^{2} x + \tan^{2} x-2 \sec x \tan x } $
$= \frac{\sec x \left( \tan x - \sec x \right)}{2 \sec^{2} x - 2 \sec x \tan x}$
$=\frac{\sec x \left(\tan x -\sec x\right)}{-2 \sec x\left(\tan x - \sec x\right)} =\frac{-1}{2}$
$\Rightarrow \frac{dy}{dx} = \frac{1.\left(\sec x
\, \times \, \tan x \,-\, \sec^{2} x\right)}{1+\left(\sec x - \tan x\right)^{2}} $
$= \frac{\sec x \left( \tan x -\sec x \right)}{1+\sec^{2} x + \tan^{2} x-2 \sec x \tan x } $
$= \frac{\sec x \left( \tan x - \sec x \right)}{2 \sec^{2} x - 2 \sec x \tan x}$
$=\frac{\sec x \left(\tan x -\sec x\right)}{-2 \sec x\left(\tan x - \sec x\right)} =\frac{-1}{2}$
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Concepts Used:
Continuity & Differentiability
Definition of Differentiability
f(x) is said to be differentiable at the point x = a, if the derivative f ‘(a) be at every point in its domain. It is given by
Definition of Continuity
Mathematically, a function is said to be continuous at a point x = a, if
It is implicit that if the left-hand limit (L.H.L), right-hand limit (R.H.L), and the value of the function at x=a exist and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is unspecified or does not exist, then we say that the function is discontinuous.