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) \[x\log a\]
E) None of the above
Correct Answer: E
Solution :
Given, \[y={{\log }_{a}}x+{{\log }_{x}}a+{{\log }_{x}}x+{{\log }_{a}}a\] \[\Rightarrow \] \[y=\frac{\log x}{\log a}-\frac{\log a}{\log x}+1+1\] On differentiating w.r.t.\[x,\]we get \[\frac{dy}{dx}=\frac{1}{x{{\log }_{a}}}-\log a{{\left( \frac{1}{\log x} \right)}^{2}}\frac{1}{x}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{1}{x\log a}-\frac{\log a}{x{{(\log x)}^{2}}}\]You need to login to perform this action.
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