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Question Number 169612 by Mastermind last updated on 04/May/22

Differentiate w.r.t ′x′   x^y +y^x =c    Mastermind

$${Differentiate}\:{w}.{r}.{t}\:'{x}'\: \\ $$$${x}^{{y}} +{y}^{{x}} ={c} \\ $$$$ \\ $$$${Mastermind} \\ $$

Answered by thfchristopher last updated on 04/May/22

⇒e^(ln x^y ) +e^(ln y^x ) =c  ⇒e^(yln x) +e^(xln y) =c  ⇒e^(yln x) ((y/x)+ln x(dy/dx))+e^(xln y) (ln y+(x/y)∙(dy/dx))=0  ⇒x^y (y/x)+x^y ln x(dy/dx)+y^x ln y+y^x (x/y)∙(dy/dx)=0  ⇒(x^y ln x+xy^(x−1) )(dy/dx)=−(x^(y−1) y+y^x ln y)  (dy/dx)=−((x^(y−1) y+y^x ln y)/(x^y ln x+xy^(x−1) ))

$$\Rightarrow{e}^{\mathrm{ln}\:{x}^{{y}} } +{e}^{\mathrm{ln}\:{y}^{{x}} } ={c} \\ $$$$\Rightarrow{e}^{{y}\mathrm{ln}\:{x}} +{e}^{{x}\mathrm{ln}\:{y}} ={c} \\ $$$$\Rightarrow{e}^{{y}\mathrm{ln}\:{x}} \left(\frac{{y}}{{x}}+\mathrm{ln}\:{x}\frac{{dy}}{{dx}}\right)+{e}^{{x}\mathrm{ln}\:{y}} \left(\mathrm{ln}\:{y}+\frac{{x}}{{y}}\centerdot\frac{{dy}}{{dx}}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}^{{y}} \frac{{y}}{{x}}+{x}^{{y}} \mathrm{ln}\:{x}\frac{{dy}}{{dx}}+{y}^{{x}} \mathrm{ln}\:{y}+{y}^{{x}} \frac{{x}}{{y}}\centerdot\frac{{dy}}{{dx}}=\mathrm{0} \\ $$$$\Rightarrow\left({x}^{{y}} \mathrm{ln}\:{x}+{xy}^{{x}−\mathrm{1}} \right)\frac{{dy}}{{dx}}=−\left({x}^{{y}−\mathrm{1}} {y}+{y}^{{x}} \mathrm{ln}\:{y}\right) \\ $$$$\frac{{dy}}{{dx}}=−\frac{{x}^{{y}−\mathrm{1}} {y}+{y}^{{x}} \mathrm{ln}\:{y}}{{x}^{{y}} \mathrm{ln}\:{x}+{xy}^{{x}−\mathrm{1}} } \\ $$

Commented by Mastermind last updated on 04/May/22

Thanks

$${Thanks} \\ $$

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