Question

If $y={log}_x\left({log}_{e}x\right)\left({log}_{e}x\right)then\frac{dy}{dx}equals$ to

• A. $\frac{1}{x{log}_x{log}_{x}x}$
• B. $\frac{1}{x{log}_{e}x}$
• C. 0
• D. none of these

Answer 1

The correct option is B $\frac{1}{x{log}_{e}x}$

Given,

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

$\frac{dy}{dx}=\frac{d}{dx}\left({\log }_x\left({\log }_e\left(x\right)\right){\log }_e\left(x\right)\right)$

${(f\cdot g)}^{\prime }=f{}^{\prime }\cdot g+f\cdot g^{\prime }\to f={\log }_x\left({\log }_e\left(x\right)\right),g={\log }_e\left(x\right)$

$=\frac{d}{dx}\left({\log }_x\left({\log }_e\left(x\right)\right)\right){\log }_e\left(x\right)+\frac{d}{dx}\left({\log }_e\left(x\right)\right){\log }_x\left({\log }_e\left(x\right)\right)$

$=\frac{1-{\log }_e\left(\log \left(x\right)\right)}{x{\log }^2\left(x\right)}{\log }_e\left(x\right)+\frac{1}{x}{\log }_x\left({\log }_e\left(x\right)\right)$

$=\frac{1}{x{\log }_e\left(x\right)}$