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Question: If x<sup>y</sup> y<sup>x</sup> = 1, then dy/dx is –...

If xy yx = 1, then dy/dx is –

A

y(y+xlog⁡y)x(ylog⁡x+x)\frac{y(y + x\log y)}{x(y\log x + x)}

B

y(x+ylog⁡x)x(y+xlog⁡y)\frac{y(x + y\log x)}{x(y + x\log y)}

C

–y(y+xlog⁡y)x(x+ylog⁡x)–\frac{y(y + x\log y)}{x(x + y\log x)}

D

None of these

Answer

–y(y+xlog⁡y)x(x+ylog⁡x)–\frac{y(y + x\log y)}{x(x + y\log x)}

Explanation

Solution

taking log

y log x + x log y = 0

yx\frac{y}{x} log x dydx+xy.dydx\frac{dy}{dx} + \frac{x}{y}.\frac{dy}{dx} + log y = 0

dydx=−[y/x+log⁡ylog⁡x+x/y]\frac{dy}{dx} = - \left\lbrack \frac{y/x + \log y}{\log x + x/y} \right\rbrack