Q. Let $fx) = \sin{x,\ g(x) = x^{2}and\ h(x) = \log_{e}{x.}}$ If $F(x) = (hogof)(x)$, then $F^{''}(x)$ is equal to
A) $a\ cosec^{3}x$
B) $2\cot{x^{2} - 4x^{2}cosec^{2}x^{2}}$
C) $2x\cot x^{2}$
D) $- 2\ cosec^{2}x$
Explanation
(d)
Given, $f(x) = \sin{x,g(x) = x^{2}}$
and $h(x) = \log_{e}x$
Also , $F(x) = (hogof)(x)$
$\therefore\ (hogof)(x) = (hog)(\sin{x)}$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ = h(\sin{x^{2})}$
$\Rightarrow F(x) = 2\log{\sin x}$
On differentiating, we get
$F^{'}(x) = 2\cot x$
Again differentiating, we get
$F^{''}(x) = - 2\ cosec^{2}x$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.