Question

Can someone help me with the partial derivative calculus problem?

Answer 1

I took a screenshot of it

Answer 2

Thx for trying!

Answer 3

I used this website to double check my partial derivative solutions
https://www.symbolab.com/solver/partial-derivative-calculator/%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20z%7D%5Cleft(%5Cfrac%7Bx-y%2Bz%7D%7Bx%2By-z%7D%5Cright)/?origin=button
but dont know the ^h works

Answer 4

the partial of w with respect to x, treat y and z as constants.

Answer 5

Assuming h stands for a fixed real number, and x,y,z are variables
$$\large{
\frac{\partial w}{\partial x} = h (\frac {x-y+z}{x+y-z})^{h-1}\cdot \frac{(x+y-z)\cdot 1 - (x-y+z)\cdot 1 }{(x+y-z)^2}\\

\frac{\partial w}{\partial y} = h (\frac {x-y+z}{x+y-z})^{h-1}\cdot \frac{(x+y-z)\cdot (-1) - (x-y+z)\cdot 1 }{(x+y-z)^2}\\

\frac{\partial w}{\partial y} = h (\frac {x-y+z}{x+y-z})^{h-1}\cdot \frac{(x+y-z)\cdot 1 - (x-y+z)\cdot (-1) }{(x+y-z)^2}

}$$

Answer 6

Thank you!

Answer 7

note that you can factor out the h* stuff expression

Answer 8

$$
\large{
x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} +z\frac{\partial w}{\partial z}
\\= x\cdot h (\frac {x-y+z}{x+y-z})^{h-1}\cdot \frac{(x+y-z)\cdot 1 - (x-y+z)\cdot 1 }{(x+y-z)^2}
\\+y\cdot h (\frac {x-y+z}{x+y-z})^{h-1}\cdot \frac{(x+y-z)\cdot (-1) - (x-y+z)\cdot 1 }{(x+y-z)^2}\\
+
z\cdot h (\frac {x-y+z}{x+y-z})^{h-1}\cdot \frac{(x+y-z)\cdot 1 - (x-y+z)\cdot (-1) }{(x+y-z)^2}

}
$$

Answer 9

Ok it makes sense know, I completely missed the fact that h is the chain rule probably because I've been solving problems all day hahah :). Do you think you could explain how you get fx+fy+fz=0

You dont have to show the work, I just dont get how these are supposed all cancel each other out.

Answer 10

@perl