If $y=\cos ^{-1} \frac{9-x^2}{9+x^2}$ then $y^{\prime}(-1)$ is equal to

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y=\cos ^{-1} \frac{9-x^2}{9+x^2}

then

y^{\prime}(-1)

is equal to

View answer and explanation

Correct option: A

-3 / 5

Option A (-3/5) is correct because after simplifying the inverse cosine function and differentiating, substituting x=-1 yields -3/5.

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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}}