If $y=A \sin \omega t$ then $\frac{d^5 y}{d t^5}=$

${A} \omega^5 \cos \left(\omega {t}-\frac{\pi}{2}\right)$

${A} \omega^5 \cos \left(\omega t+\frac{\pi}{2}\right)$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

y=A \sin \omega t

then

\frac{d^5 y}{d t^5}=

View answer and explanation

Correct option: D

{A} \omega^5 \sin \left(\omega {t}+\frac{\pi}{2}\right)

Option D (

{A} \omega^5 \sin \left(\omega {t}+\frac{\pi}{2}\right)

) is correct because it represents the fifth derivative of the given function

y=A \sin \omega t

, obtained by repeatedly differentiating and using trigonometric identities.

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}}