Question

If $f\left(x\right)$ and $g\left(x\right)$ are two functions with $g\left(x\right)=x-\frac{1}{x}$ and $f\circ g\left(x\right)=x^3-\frac{1}{x^3}$, then $f^{{}^{\prime }}\left(x\right)$ is equal to

• A. $3x^2+3$
• B. $x^2-\frac{1}{x^2}$
• C. $1+\frac{1}{x^2}$
• D. $3x^2+\frac{3}{x^4}$

Answer 1

The correct option is A $3x^2+3$

$f\circ g\left(x\right)=x^3-\frac{1}{x^3}$
Writing $x^3-\frac{1}{x^3}$ using ${(a-b)}^3=a^3-b^3-3ab(a-b)$, we have
${(x-\frac{1}{x})}^3=x^3-\frac{1}{x^3}-3x\cdot \frac{1}{x}(x-\frac{1}{x})$
$\Rightarrow x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$
We have,
$f\left(g\left(x\right)\right)=x^3-\frac{1}{x^3}={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$
As $g\left(x\right)=x-\frac{1}{x}$, this yields
$f(x-\frac{1}{x})={(x-\frac{1}{x})}^3+3(x-\frac{1}{x})$
On putting $x-\frac{1}{x}=t$, we get
$f\left(t\right)=t^3+3t$
Thus, $f\left(x\right)=x^3+3x$
and $f^{{}^{\prime }}\left(x\right)=3x^2+3$