Let $f\left( x \right) = \sin x,{\rm{g}}\left( x \right) = {x^2}$ and $h\left( x \right) = {\log _e}x.$ If $F\left( x \right) = \left( {ho{\rm{g}}\,\,of} \right)\left( x \right),$ then $F''\left( x \right)$ is equal to

$ - 2\,{\rm{cose}}{{\rm{c}}^2}x$

$a\,\,{\rm{cose}}{{\rm{c}}^3}x$

$2\cot {x^2} - 4{x^2}{\rm{cose}}{{\rm{c}}^2}{x^2}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

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Application +4 | -1

Let

f\left( x \right) = \sin x,\mathrm{g}\left( x \right) = {x^2}

and

h\left( x \right) = {\log _e}x.

F\left( x \right) = \left( {ho\mathrm{g}\,\,of} \right)\left( x \right),

then

F''\left( x \right)

is equal to

View answer and explanation

Correct option: D

- 2\,\mathrm{cose}{\mathrm{c}^2}x

Option D (

- 2\,\mathrm{cose}{\mathrm{c}^2}x

) is correct because it represents the second derivative of the composite function F(x) after applying the chain rule and trigonometric identities.