Question

If $y={\log }_a\left(x\right),x>0$, then $\frac{dy}{dx}=$

• A. $\frac{1}{x}$
• B. $\frac{1}{ax}$
• C. $\frac{1}{x}{\log }_x\left(a\right)$
• D. $\frac{1}{x}{\log }_a\left(e\right)$

Answer 1

The correct option is D

$\frac{1}{x}{\log }_a\left(e\right)$

Explanation for the correct option:

$y={\log }_a\left(x\right)$

$\Rightarrow$ $y=\frac{\log \left(x\right)}{\log \left(a\right)}$ $\left[\because {\log }_n\left(m\right)=\frac{\log \left(m\right)}{\log \left(n\right)}\right]$

Differentiating with respect to $x$ we get,

$\Rightarrow$$\frac{dy}{dx}=\frac{1}{x}\times \frac{1}{\log \left(a\right)}$ $\left[\because \frac{d}{\mathrm{dx}}\left(\log \left(x\right)\right)=\frac{1}{x}\right]$

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{x}\times \frac{\log \left(e\right)}{\log \left(a\right)}$ $\left[\because {\log }_e\left(e\right)=1\right]$

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{x}\times {\log }_a\left(e\right)$ $\left[\because {\log }_n\left(m\right)=\frac{\mathrm{logm}}{\mathrm{logn}}\right]$

$\Rightarrow$ $\frac{dy}{dx}=\frac{1}{x}\left({\log }_a\left(e\right)\right)$

Hence, option(D) i.e. $\frac{1}{x}{\log }_a\left(e\right)$ is the correct answer.