If $x^{y}=e^{x-y}$, show that $\dfrac{dy}{dx}=\dfrac{\log x}{\{\log(xe)\}^{2}}$.

CBSE Class 12 · 2011 · DifferentiationLong Answer

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

, show that

\frac{d y}{d x} = \frac{lo g x}{{ lo g ( x e ) } ^{2}}

Answer: Proved

Take logarithm of both sides:

y lo g x = (x - y) lo g e = x - y

y (1 + lo g x) = x

, giving

y = \frac{x}{1 + lo g x}

.
Differentiate by the quotient rule (note

\frac{d}{d x} (1 + lo g x) = \frac{1}{x}

\frac{d y}{d x} = \frac{( 1 + lo g x ) \cdot 1 - x \cdot \frac{1}{x}}{( 1 + lo g x ) ^{2}} = \frac{lo g x}{( 1 + lo g x ) ^{2}}

Since

1 + lo g x = lo g e + lo g x = lo g (x e)

\frac{d y}{d x} = \frac{lo g x}{{ lo g ( x e ) } ^{2}}

CBSE Class 12 Board · 2011 · Subjective