If ${x^y} = {e^{x - y}}$, then $\frac{{dy}}{{dx}}$ is equal to

${\left( {1 + \log x} \right)^{ - 1}}$

JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

{x^y} = {e^{x - y}}

, then

\frac{{dy}}{{dx}}

is equal to

View answer and explanation

Correct option: C

\log x.{\left( {1 + \log x} \right)^{ - 2}}

Option C,

\log x.{\left( {1 + \log x} \right)^{ - 2}}

, is correct because it represents the derivative dy/dx obtained by implicit differentiation and simplification of the given equation

{x^y} = {e^{x - y}}

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