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Question Number 60007 by meme last updated on 17/May/19

$$dfoff\left(x\right)=x^x{y}^y$$

Answered by alex041103 last updated on 18/May/19

$$ifyoumeanf(x,y)=x^x{y}^{y}then$$$$df=\frac{\partial f}{\partial x}dx+\frac{\partial f}{\partial y}dy=$$$$=y^y\frac{\partial }{\partial x}\left(x^x\right)dx+x^x\frac{\partial }{\partial y}\left(y^y\right)dy=$$$$Ingeneral{let}^{\prime }sfind\frac{\partial }{\partial z}{z}^z:$$$$z^z=e^{zln\left(z\right)}\Rightarrow \frac{\partial }{\partial z}\left(z^z\right)=\frac{\partial }{\partial z}\left(e^{zln\left(z\right)}\right)=$$$$=\frac{\partial }{\partial \left(zln\left(z\right)\right)}\left(e^{zln\left(z\right)}\right)\times \frac{\partial \left(zln\left(z\right)\right)}{\partial z}=$$$$=e^{zln\left(z\right)}\times (1\times ln\left(z\right)+z\times \frac{1}{z})=$$$$=z^z(1+ln\left(z\right))=\frac{\partial \left(z^z\right)}{\partial z}$$$$\Rightarrow df=x^x{y}^y(1+ln\left(x\right))dx+x^x{y}^y(1+ln\left(y\right))dy$$$$\Rightarrow df=f[(1+ln\left(x\right))dx+(1+ln\left(y\right))dy]$$$$youcangoalittlebitfurther:$$$$\frac{df}{f}=dx+ln\left(x\right)dx+dy+ln\left(y\right)dy=$$$$=d(x+y)+ln\left(x^{dx}\right)+ln\left(y^{dy}\right)=$$$$=d(x+y)+ln\left(x^{dx}{y}^{dy}\right)$$$$\Rightarrow df=x^x{y}^y[d(x+y)+ln\left(x^{dx}{y}^{dy}\right)]$$

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