JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

f(x)=\sin x, g(x)=x^{2}, h(x)=\log x

and

F(x)=(

hogof)

(x)

, then

F^{\prime \prime}(x)

View answer and explanation

Correct option: D

-2 \operatorname{cosec}^{2} x

The correct answer is D (

-2 \operatorname{cosec}^{2} x

) because it is the second derivative of the function F(x) after simplification and differentiation.

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

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x

y

x>0

x y=1

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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}}