Differentiation Question 246
Question: If $ y={\log _{10}}x+{\log _{x}}10+{\log _{x}}x+{\log _{10}}10, $ then $ \frac{dy}{dx}= $
Options:
A) $ \frac{1}{x{\log _{e}}10}-\frac{{\log _{e}}10}{x{{({\log _{e}}x)}^{2}}} $
B) $ \frac{1}{x{\log _{e}}10}-\frac{1}{x{\log _{10}}e} $
C) $ \frac{1}{x{\log _{e}}10}-\frac{{\log _{e}}10}{x{{({\log _{e}}x)}^{2}}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ y={\log _{10}}x+{\log _{x}}10+{\log _{x}}x+{\log _{10}}10 $
$ ={\log _{10}}x+\frac{{\log _{e}}10}{\log x}+1+1 $
$ \Rightarrow \frac{dy}{dx}=\frac{1}{x}{\log _{10}}e-\frac{{\log _{e}}10}{x{{({\log _{e}}x)}^{2}}} $ .
BETA
