Question

Let $f$ be a differentiable function defined for all $x\in R$ such that $f\left(x^3\right)=x^5$ for all $x\in R,x\ne 0.$ Then, the value of $f^{\prime }\left(8\right)$ is

• A. $20$
• B. $\frac{20}{3}$
• C. $\frac{5}{3}$
• D. none of these

Answer 1

Answer: (B)

$\frac{20}{3}$

$f\left(x^3\right)=x^5$ for all $x\in R,x\ne 0$
Differentiating w.r. to $x$
$\frac{d}{dx}f\left(x^3\right)=\frac{d}{dx}{x}^5$
$f^{{}^{\prime }}\left(x^3\right)\frac{d}{dx}{x}^3=5x^4$
$f^{{}^{\prime }}\left(x^3\right)=\frac{5}{3}{x}^2$
Putting $x^3=8$ to find $f^{{}^{\prime }}\left(8\right)$
$\therefore f^{{}^{\prime }}\left(8\right)=\frac{5.{\left(2\right)}^2}{3}=\frac{20}{3}$