Answer to Question #254386 in Calculus for Hasni

Question #254386

Let x>0, and y be a positive function of x such that x^y=y^-x.

Find y′



1
Expert's answer
2021-10-21T14:45:24-0400

xy=y(x)x^{y}=y^{(-x)}

Taking logerithm with respect to base e on both sides

lnxy=lny(x)\ln{x^{y}}=\ln{y^{(-x)}}

=> ylnx=xlnyy\ln{x}=-x\ln{y}

Differentiating with respect to x

d(ylnx)dx=d(xlny)dx\frac{d(y\ln{x})}{dx}= - \frac{d(x\ln{y})}{dx}

=> yddx(lnx)+lnxdydx=xddx(lny)lnyddx(x)y\frac{d}{dx} (\ln{x})+\ln{x} \frac{dy}{dx}=-x\frac{d}{dx}(\ln{y})-\ln{y} \frac{d}{dx}{(x)}

=> y.1x+lnxdydx=x.1y.dydxlnyy.\frac{1}{x} + ln{x} \frac{dy}{dx}=-x.\frac{1}{y}.\frac{dy}{dx}-\ln{y}

=> lnxdydx+x.1y.dydx\ln{x} \frac{dy}{dx}+x.\frac{1}{y}.\frac{dy}{dx} = yxlny-\frac{y}{x} - \ln{y}

=> x(ylnx+x)dydx=y(y+xlny)x(y\ln{x}+x)\frac{dy}{dx}= -y(y+x\ln{y})

=>

=> dydx=\frac{dy}{dx}= y(y+xlny)x(ylnx+x)-\frac{ y(y+x\ln{y})}{x(y\ln{x}+x)}

So , y' = y(y+xlny)x(ylnx+x)-\frac{ y(y+x\ln{y})}{x(y\ln{x}+x)}





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