Q. If $f(x)$ and g$\ (x)$ are two functions with g$\ (x) = x - \frac{1}{x}$ and $fog\ (x) = x^{3} - \frac{1}{x^{3}},$ then $f^{'}(x)$ is
A) $3x^{2} + 3$
B) $x^{2} - \frac{1}{x^{2}}$
C) $1 + \frac{1}{x^{2}}$
D) $3x^{2} + \frac{3}{x^{4}}$
Explanation
(a)
Since, $f\left( x - \frac{1}{x} \right) = x^{3} - \frac{1}{x^{3}}$
$f\left( x - \frac{1}{x} \right) = \left( x - \frac{1}{x} \right)\left( x^{2} + \frac{1}{x^{2}} + 1 \right)$
$= \left( x - \frac{1}{x} \right)$ $\ \left\lbrack \left( x - \frac{1}{x} \right)^{2} + 3 \right\rbrack$
$\Rightarrow f(x) = x\left( x^{2} + 3 \right) = x^{3} + 3x\$
$\Rightarrow \ f^{'}(x) = \ 3x^{2} + 3$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.