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Question Number 82993 by aseer imad last updated on 26/Feb/20

Answered by TANMAY PANACEA last updated on 27/Feb/20

$${(x+y)}^{x+y}=x^y+y^x$$$$a=b+c$$$$\frac{da}{dx}=\frac{db}{dx}+\frac{dc}{dx}$$$$nowa={(x+y)}^{x+y}$$$$lna=(x+y)ln(x+y)$$$$\frac{1}{a}\times \frac{da}{dx}=(x+y)\times \frac{1}{x+y}(1+\frac{dy}{dx})+ln(x+y)\times (1+\frac{dy}{dx})$$$$\frac{1}{a}\times \frac{da}{dx}=(1+\frac{dy}{dx})\{1+ln(x+y)\}$$$$\frac{da}{dx}={(x+y)}^{x+y}\times (1+\frac{dy}{dx})\{1+ln(x+y)\}$$$$b=x^y$$$$lnb=ylnx$$$$\frac{1}{b}\times \frac{db}{dx}=\frac{y}{x}+\frac{dy}{dx}lnx$$$$\frac{db}{dx}=x^y(\frac{y}{x}+\frac{dy}{dx}lnx)$$$$c=y^x$$$$lnc=xlny$$$$\frac{1}{c}\frac{dc}{dx}=\frac{x}{y}\frac{dy}{dx}+lny$$$$\frac{dc}{dx}=y^x(\frac{x}{y}\frac{dy}{dx}+lny)$$$$\frac{da}{dx}=\frac{db}{dx}+\frac{dc}{dx}$$$${(x+y)}^{x+y}(1+\frac{dy}{dx})\{1+ln(x+y)\}=x^y(\frac{y}{x}+\frac{dy}{dx}lnx)+y^x(\frac{x}{y}\frac{dy}{dx}+lny)$$$$nowsimplify...$$$$$$

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