Question

A differentiable function $\mathrm{f}\left(\mathrm{x}\right)$ is defined for all$x>0$, and satisfies $\mathrm{f}\left({\mathrm{x}}^3\right)=4{\mathrm{x}}^4$, for all$x>0$. The value of$\mathrm{f}’\left(8\right)$ is

• A. $\frac{16}{3}$
• B. $\frac{32}{3}$
• C. $\frac{16\sqrt{2}}{3}$
• D. $\frac{32\sqrt{2}}{3}$

Answer 1

The correct option is B

$\frac{32}{3}$

Explanation for the correct option:

Step1. Differentiate the given function:

$\begin{array}{rcl}\mathrm{f}\left({\mathrm{x}}^3\right)& =& 4{\mathrm{x}}^4,\mathrm{for}\mathrm{all}\mathrm{x}>0\\ & \Rightarrow & {\mathrm{f}}^{\text{'}}\left({\mathrm{x}}^3\right)3{\mathrm{x}}^2=16{\mathrm{x}}^3\left[\because Applyf\left(x\right)=x^n\Rightarrow f\text{'}\left(x\right)=n{x}^{n-1}\right]\\ & \Rightarrow & \mathrm{f}\text{'}\left({\mathrm{x}}^3\right)=\frac{16{\mathrm{x}}^3}{3{\mathrm{x}}^2}\\ & \Rightarrow & \mathrm{f}\text{'}\left({\mathrm{x}}^3\right)=\frac{16\mathrm{x}}{3}\end{array}$

Step2. Calculate the value of ${\mathbf{f}}^{\mathbf{\text{'}}}\left(8\right)$:

$\begin{array}{rcl}{\mathrm{x}}^3& =& 8\mathbf{[}\mathbf{C}\mathbf{o}\mathbf{m}\mathbf{p}\mathbf{a}\mathbf{r}\mathbf{e}\mathbf{}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{}\mathbf{f}\mathbf{u}\mathbf{n}\mathbf{c}\mathbf{t}\mathbf{i}\mathbf{o}\mathbf{n}\mathbf{s}\mathbf{]}\\ & \Rightarrow & \mathrm{x}=2\\ & \Rightarrow & {\mathrm{f}}^{\text{'}}\left(8\right)=\frac{16\left(2\right)}{3}\mathbf{[}\mathbf{S}\mathbf{u}\mathbf{b}\mathbf{s}\mathbf{t}\mathbf{i}\mathbf{t}\mathbf{u}\mathbf{t}\mathbf{e}\mathbf{}\mathbf{t}\mathbf{h}\mathbf{e}\mathbf{}\mathbf{v}\mathbf{a}\mathbf{l}\mathbf{u}\mathbf{e}\mathbf{}\mathbf{o}\mathbf{f}\mathbf{}\mathbf{x}\mathbf{=}\mathbf{2}\mathbf{}\mathbf{i}\mathbf{n}\mathbf{}\mathbf{f}\mathbf{\text{'}}\mathbf{\left(}{\mathbf{x}}^{\mathbf{3}}\mathbf{\right)}\mathbf{]}\\ & =& \frac{32}{3}\end{array}$

Hence, Option(B) is the correct answer.