Nice derivative

Calculus Level pending

If x y = e x − y x^y=e^{x-y} , then what is the value of d y d x ? \frac{dy}{dx} ?

ln ⁡ x ( 1 + ln ⁡ x ) 2 \frac{\ln x}{(1+\ln x)^2} ln ⁡ x ( 1 + ln ⁡ x ) \frac{\ln x}{(1+\ln x)} x ( 1 + ln ⁡ x ) 2 \frac{x}{(1+\ln x)^2} ln ⁡ x ( 1 − ln ⁡ x ) 2 \frac{\ln x}{(1-\ln x)^2}

Hana Wehbi by Hana Wehbi , 51, USA

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1 solution

Hana Wehbi
Mar 23, 2017

x y = e x − y \large x^y=e^{x-y}

Take logarithms base e e on both sides:

Therefore y ln ⁡ x = ln ⁡ ( e x − y ) \large y \ln x = \ln(e^{x-y})

⟹ y ln ⁡ x = x − y \large \implies y \ln x= x-y

⟹ y ( 1 + ln ⁡ x ) = x \large \implies y(1+\ln x) = x

y = x 1 + ln ⁡ x \large y = \frac{x}{1+\ln x}

This differentiated using the quotient rule: 1 + ln ⁡ x − 1 x ( x ) ( 1 + ln ⁡ x ) 2 = ln ⁡ x ( 1 + ln ⁡ x ) 2 \large\frac{1 + \ln x - \frac{1}{x}( x)}{(1+\ln x)^2} = \boxed{ \frac {\ln x}{(1+\ln x)^2}}

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