Question

Suppose that $f\left(x\right)$ is a differentiable function such that $f\text{'}\left(x\right)$ is continuous $f\text{'}\left(0\right)=1$ and $f\text{'}\text{'}\left(0\right)$ does not exist. If $g\left(x\right)=xf\text{'}\left(x\right)$. Then,

• A. $g\text{'}\left(0\right)$ does not exist
• B. $g\text{'}\left(0\right)=0$
• C. $g\text{'}\left(0\right)=1$
• D. $g\text{'}\left(0\right)=2$

Answer 1

The correct option is C

$g\text{'}\left(0\right)=1$

Explanation for the correct option:

Given:

$f\text{'}\left(0\right)=1$

$g\left(x\right)=xf\text{'}\left(x\right)$

By Differentiating it, we get

$g\text{'}\left(x\right)=f\text{'}\left(x\right)+xf\text{'}\text{'}\left(x\right)$ $\left[\mathbf{\because }\frac{\mathbf{d}}{\mathbf{d}\mathbf{x}}\left(uv\right)\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\frac{\mathbf{d}\mathbf{v}}{\mathbf{d}\mathbf{x}}\mathbf{+}\mathit{v}\frac{\mathbf{d}\mathbf{u}}{\mathbf{d}\mathbf{x}}\right]$

$\Rightarrow g\text{'}\left(0\right)=f\text{'}\left(0\right)+0f\text{'}\text{'}\left(0\right)$

$\therefore$$g\text{'}\left(0\right)=\mathbf{1}$

Hence, Option ‘C’ is Correct.