A) \[\log x.{{[\log (ex)]}^{-2}}\]
B) \[\log x.{{[\log (ex)]}^{2}}\]
C) \[\log x.{{(\log x)}^{2}}\]
D) None of these
Correct Answer: A
Solution :
\[{{x}^{y}}={{e}^{x-y}}\] Þ \[y\log x=x-y\] Þ \[y=\frac{x}{1+\log x}\] Þ \[\frac{dy}{dx}=\log x{{(1+\log x)}^{-2}}=\log x{{[\log ex]}^{-2}}\].You need to login to perform this action.
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