Question

The derivative of $f\left(x\right)=\left|x^3\right|$ at$x=0$ is

• A. $0$
• B. 1$0$
• C. $-1$
• D. not defined

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Finding the derivative of the given function:

$$\begin{gathered} f\left(x\right)\mathit{}\mathit{=}\mathit{}\mathit{\left|}{\mathit{x}}^{\mathit{3}}\mathit{\right|} \\ f\mathit{\left(}\mathit{x}\mathit{\right)}\mathit{}\mathit{=}\mathit{}{x}^{\mathit{3}}\mathit{}\mathit{,}\mathit{}\mathit{}x\mathit{\ge }o \\ f\mathit{\left(}x\mathit{\right)}\mathit{}\mathit{=}\mathit{-}{x}^{\mathit{3}}\mathit{}\mathit{,}\mathit{}\mathit{}x\mathit{<}o \end{gathered}$$

Using the formula$\mathit{f}\left(x\right)\mathbf{=}\underset{\mathbf{h}\mathbf{\to }\mathbf{x}\mathbf{\pm }}{\mathbf{l}\mathbf{i}\mathbf{m}}\left(\frac{f\left(x-h\right)-f\left(x\right)}{\left(x-h\right)-x}\right)$

Now LHD at $x\mathit{=}o$

$$\begin{gathered} =\underset{h\to 0-}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{0-h-0} \\ =\underset{h\to 0-}{\lim }\frac{f\left(-h\right)-f\left(0\right)}{-h} \\ =\underset{h\to 0-}{\lim }\frac{-{\left(-h\right)}^3-0}{-h} \\ =\underset{h\to 0-}{\lim }-h^2 \\ =0 \end{gathered}$$

Therefore LHD$=0$

Now, RHD at $x\mathit{=}o$

$$\begin{gathered} =\underset{h\to 0+}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{0+h-0} \\ =\underset{h\to 0+}{\lim }\frac{f\left(h\right)-0}{h} \\ =\underset{h\to 0+}{\lim }\frac{h^3-0}{h} \\ =\underset{h\to 0+}{\lim }{h}^2 \\ =0 \\ \mathrm{Therefore},\mathrm{LHD}=\mathrm{RHD}=0 \\ f\text{'}\left(0\right)=0 \end{gathered}$$

Therefore, the correct answer is option (A).