Question

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

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

Answer 1

The correct option is A

$3x^2+3$

Explanation for the correct option.

Find the value of $f\text{'}\left(x\right)$:

$\begin{array}{rcl}fog\left(x\right)& =& x^3-\frac{1}{x^3}\\ & \Rightarrow & f\left(x-\frac{1}{x}\right)=\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)\left[by{a}^3-b^3=\left(a-b\right)\left(a^2+b^2+ab\right)\right]\\ & =& \left(x-\frac{1}{x}\right)\left[{\left(x-\frac{1}{x}\right)}^2+2+1\right]\left[by{a}^2+b^2={\left(a-b\right)}^2-2ab\right]\\ & =& \left(x-\frac{1}{x}\right)\left[{\left(x-\frac{1}{x}\right)}^2+3\right]\end{array}$

Let $x-\frac{1}{x}=t$

$\begin{array}{rcl}f\left(t\right)& =& t^3+3t\\ \Rightarrow f\left(x\right)& =& x^3+3x\\ & \Rightarrow & f\text{'}\left(x\right)=3x^2+3\end{array}$

Hence, option A is correct.