If $ y=\sqrt{\frac{1+\tan x}{1-\tan x}} $, then $ \frac{d y}{d x} $ is equal to

$ \frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \sec \left(\frac{\pi}{4}+x\right) $

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

y=\sqrt{\frac{1+\tan x}{1-\tan x}}

, then

\frac{d y}{d x}

is equal to

View answer and explanation

Correct option: A

\frac{1}{2} \sqrt{\frac{1-\tan x}{1+\tan x}} \cdot \sec ^{2}\left(\frac{\pi}{4}+x\right)

Option A is correct because the derivative is found using the chain rule and trigonometric identities to simplify the expression.

Practice All Chapter MCQs →

f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

x>0

x y=1

x+y

f(x)=[x] \sin (\pi x)

[x]

x

x=k, k

f(x)

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

\frac{{dy}}{{dx}}

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

z = y + f\left( v \right),

v = \left( {\frac{x}{y}} \right),

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}