A) \[\frac{1}{x}+x\log a\]
B) \[\frac{\log a}{x}+\frac{x}{\log a}\]
C) \[\frac{1}{x\log a}+x\log a\]
D) \[\frac{1}{x\log a}-\frac{\log a}{x{{(\log x)}^{2}}}\]
E) none of these
Correct Answer: D
Solution :
\[y={{\log }_{a}}+\frac{\log a}{\log x}+1+1\] On differentiating, we get \[\frac{dy}{dx}=\frac{1}{x}{{\log }_{a}}e-\log a{{\left( \frac{1}{\log x} \right)}^{2}}.\frac{1}{x}\] \[=\frac{1}{x\log a}-\frac{\log a}{x{{(\log x)}^{2}}}\]
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