df-of-f-x-x-x-y-y-

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

d f o f f (x) = x^{x} y^{y}

Answered by alex041103 last updated on 18/May/19

i f y o u m e a n f (x, y) = x^{x} y^{y} t h e n

d f = \frac{\partial f}{\partial x} d x + \frac{\partial f}{\partial y} d y =

= y^{y} \frac{\partial}{\partial x} (x^{x}) d x + x^{x} \frac{\partial}{\partial y} (y^{y}) d y =

I n g e n e r a l {l e t}^{'} s f i n d \frac{\partial}{\partial z} z^{z} :

z^{z} = e^{z l n (z)} \Rightarrow \frac{\partial}{\partial z} (z^{z}) = \frac{\partial}{\partial z} (e^{z l n (z)}) =

= \frac{\partial}{\partial (z l n (z))} (e^{z l n (z)}) \times \frac{\partial (z l n (z))}{\partial z} =

= e^{z l n (z)} \times (1 \times l n (z) + z \times \frac{1}{z}) =

= z^{z} (1 + l n (z)) = \frac{\partial (z^{z})}{\partial z}

\Rightarrow d f = x^{x} y^{y} (1 + l n (x)) d x + x^{x} y^{y} (1 + l n (y)) d y

\Rightarrow d f = f [(1 + l n (x)) d x + (1 + l n (y)) d y]

y o u c a n g o a l i t t l e b i t f u r t h e r :

\frac{d f}{f} = d x + l n (x) d x + d y + l n (y) d y =

= d (x + y) + l n (x^{d x}) + l n (y^{d y}) =

= d (x + y) + l n (x^{d x} y^{d y})

\Rightarrow d f = x^{x} y^{y} [d (x + y) + l n (x^{d x} y^{d y})]