Question

xy=ex-y(differentiate following by using logathimic method)

Answer 1

AnSwEr :

$\rightarrow \sf{xy \: = \: e^{x \: - \: y}}$

Take log both sides

$\rightarrow \sf{\log (xy) \: = \: \log e^{x \: - \: y}} \\ \\ \rightarrow \sf{\log (xy) \: = \: (x \: - \: y) \log e} \\ \\ \rightarrow \sf{\log x \: + \: \log y \: = \: x \: - \: y}$

Differentiate w.r.t x

$\rightarrow \sf{\dfrac{d(\log x \: + \: \log y)}{dx} \: = \: \dfrac{d(x \: - \: y)}{dx}} \\ \\ \rightarrow \sf{\dfrac{d(\log x) }{dx} \: + \: \dfrac{d(\log y}{dx} \: = \: \dfrac{dx}{dx} \: - \: \dfrac{dy}{dx}} \\ \\ \rightarrow \sf{\dfrac{1}{x} \: + \: \dfrac{d(\log y) }{dy} . \dfrac{dy}{dx} \: = \: 1 \: - \: \dfrac{dy}{dx}} \\ \\ \rightarrow \sf{\dfrac{1}{x} \: + \: \dfrac{1}{y} . \dfrac{dy}{dx} \: = \: 1 \: - \: \dfrac{dy}{dx}} \\ \\ \rightarrow \sf{\dfrac{1}{dy} . \dfrac{dy}{dx} \: + \: \dfrac{dy}{dx} \: = \: 1 \: - \: \dfrac{1}{x}} \\ \\ \rightarrow \sf{\dfrac{dy}{dx} \bigg( \dfrac{1}{y} \: + \: 1 \bigg) \: = \: \bigg( 1 \: - \: \dfrac{1}{x} \bigg)} \\ \\ \rightarrow \sf{\dfrac{dy}{dx} \bigg( \dfrac{1 \: + \: y}{y} \bigg) \: = \: \bigg( \dfrac{x \: - \: 1}{x} \bigg) } \\ \\ \rightarrow \sf{\dfrac{dx}{dy} \: = \: \bigg( \dfrac{x \: - \: 1}{y} \bigg) \bigg( \dfrac{y}{1 \: + \: y} \bigg) } \\ \\ \rightarrow \sf{\dfrac{dy}{dx} \: = \: \dfrac{y(x \: - \: 1)}{x (1 \: + \: y)}}$