If $ x=3 \tan t $ and $ y=3 \sec t $, then the value of $ \frac{d^{2} y}{\mathrm{dx}^{2}} $ at $ \mathrm{t}=\frac{\pi}{4} $, is:

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

x=3 \tan t

and

y=3 \sec t

, then the value of

\frac{d^{2} y}{\mathrm{dx}^{2}}

\mathrm{t}=\frac{\pi}{4}

, is:

View answer and explanation

Correct option: B

\frac{1}{6 \sqrt{2}}

Option B (

\frac{1}{6 \sqrt{2}}

) is correct because it is the value of the second derivative calculated using parametric differentiation at the given point.

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}}