If variables $x$ and $y$ are related by the equation $x = \mathop \smallint \limits_0^y \frac{1}{{\sqrt {1 + 9{u^2}} }}du$, then $\frac{{dy}}{{dx}}$ is equal to

$\frac{1}{{\sqrt {1 + 9{y^2}} }}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

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Application +4 | -1

If variables

x

and

y

are related by the equation

x = \int \limits_0^y \frac{1}{{\sqrt {1 + 9{u^2}} }}du

, then

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

is equal to

View answer and explanation

Correct option: B

\sqrt {1 + 9{y^2}}

Option B (

\sqrt {1 + 9{y^2}}

) is correct because it is the direct result of differentiating the inverse function and applying the Fundamental Theorem of Calculus.

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}}