Question:
If $y=\log \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) ,$ then $ \frac{dy}{dx} = $
If $y=\log \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) ,$ then $ \frac{dy}{dx} = $
Updated On: Jun 21, 2022
- $\sec \: x$
- $\sin\: x$
- $cosec \: x$
- $\sec \: \frac{x}{2}$
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The Correct Option is A
Solution and Explanation
Given, $y=\log \tan\left(\frac{\pi}{4} + \frac{x}{2}\right) $
$\Rightarrow \frac{dy}{dx} = \frac{1}{\tan \left(\frac{\pi }{4} + \frac{x}{2}\right)} \frac{1}{2} \sec^{2} \left(\frac{\pi }{4} + \frac{x}{2}\right) $
$= \frac{1}{2} \left[ \frac{1}{\tan \left(\frac{\pi }{4} + \frac{x}{2}\right)} + \frac{\tan^{2} \left(\frac{\pi }{4} + \frac{x}{2}\right)}{\tan \left(\frac{\pi }{4} + \frac{x}{2}\right)}\right] = $
$= \frac{1}{2} \left[ \frac{\cos \left(\frac{\pi }{4} + \frac{x}{2}\right)}{\sin \left(\frac{\pi }{4} + \frac{x}{2}\right)} + \frac{\sin \left(\frac{\pi }{4} + \frac{x}{2}\right)}{\cos \left(\frac{\pi }{4} + \frac{x}{2}\right)}\right] $
$\frac{dy}{dx} = \frac{1}{\sin \left( \frac{\pi}{2} + x\right)} \left[ \because \:\: \cos^{2} x \sin^{2} x = 1\right] $
$= \frac{1}{\cos x} =\sec x \left[ \because \:\: \sin \left( \frac{\pi }{2} + \theta\right) =\cos \theta\right]$
$\Rightarrow \frac{dy}{dx} = \frac{1}{\tan \left(\frac{\pi }{4} + \frac{x}{2}\right)} \frac{1}{2} \sec^{2} \left(\frac{\pi }{4} + \frac{x}{2}\right) $
$= \frac{1}{2} \left[ \frac{1}{\tan \left(\frac{\pi }{4} + \frac{x}{2}\right)} + \frac{\tan^{2} \left(\frac{\pi }{4} + \frac{x}{2}\right)}{\tan \left(\frac{\pi }{4} + \frac{x}{2}\right)}\right] = $
$= \frac{1}{2} \left[ \frac{\cos \left(\frac{\pi }{4} + \frac{x}{2}\right)}{\sin \left(\frac{\pi }{4} + \frac{x}{2}\right)} + \frac{\sin \left(\frac{\pi }{4} + \frac{x}{2}\right)}{\cos \left(\frac{\pi }{4} + \frac{x}{2}\right)}\right] $
$\frac{dy}{dx} = \frac{1}{\sin \left( \frac{\pi}{2} + x\right)} \left[ \because \:\: \cos^{2} x \sin^{2} x = 1\right] $
$= \frac{1}{\cos x} =\sec x \left[ \because \:\: \sin \left( \frac{\pi }{2} + \theta\right) =\cos \theta\right]$
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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.