JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Application +4 | -1

\frac{{{d^2}}}{{d{x^2}}}(2\cos x\cos 3x)

is equal to

View answer and explanation

Correct option: D

- {2^2}(\cos 2x + {2^2}\cos 4x)

Option D (

- {2^2}(\cos 2x + {2^2}\cos 4x)

) is correct because it is the second derivative of the given function, calculated using trigonometric identities and differentiation.

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f(x)=|x|^{|\tan x|}

f^{\prime}\left(\frac{-\pi}{6}\right)

x

y

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x y=1

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[x]

x

x=k, k

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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),

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v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}