If $y=\sin ^{-1} \frac{\sqrt{(1+x)}+\sqrt{(1-x)}}{2}$, then $\frac{d y}{d x}=$

$\frac{1}{\sqrt{\left(1-x^2\right)}}$

$-\frac{1}{\sqrt{\left(1-x^2\right)}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

y=\sin ^{-1} \frac{\sqrt{(1+x)}+\sqrt{(1-x)}}{2}

, then

\frac{d y}{d x}=

View answer and explanation

Correct option: C

-\frac{1}{2 \sqrt{\left(1-x^2\right)}}

Option C (

-\frac{1}{2 \sqrt{\left(1-x^2\right)}}

) is correct because it is the result of simplifying the expression using trigonometric substitution and then differentiating.

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}}