Hello, I have the next exercise about chain rule in multivariable calculus:

I have to show that the differential equation:
y(dz/dx)-x(dz/dy) = (y-x)z
can be change to the equation:
(dw/dv) = 0
using the new variables:
u= (x^2)+(y^2)
v = (1/x)+(1/y)
w = ln(z)-(x+y)
For that I have to do some steps,in the first one I have to find the derivative of w respect to v
My problem is that I don't know what to do in order to express that derivative of w respect to v
Please could anyone help me giving me some steps o what to do?
I would really appreaciate the help

dw/dv = (dw/dx) / (dv/dx)

dw/dx = ∂w/∂x + ∂w/∂y dy/dx + ∂w/∂z dz/dx

and similarly for u,v

See where you get with that.

To find the derivative of w with respect to v, we will use the chain rule. The chain rule states that if we have a composite function u(v) = f(g(v)), then the derivative of u with respect to v can be found by multiplying the derivative of f with respect to g, multiplied by the derivative of g with respect to v.

In our case, we have w = ln(z) - (x + y), where z = z(x, y), x = x(v), and y = y(v). So, we need to find the derivative of w with respect to v, which can be expressed as dw/dv. Let's break it down step by step:

Step 1: Find the partial derivatives of w with respect to x, y, and z.

dw/dx = d/dx (ln(z) - (x + y))
= -1 - 1 (since the derivative of x is 1 and the derivative of y is 1)
= -2

dw/dy = d/dy (ln(z) - (x + y))
= -1 - 1 (since the derivative of x is 0 and the derivative of y is 1)
= -2

dw/dz = d/dz (ln(z) - (x + y))
= 1/z - 0 (since the derivative of ln(z) is 1/z and the derivative of (x + y) is 0)
= 1/z

Step 2: Find the partial derivatives of x and y with respect to v.

To find dx/dv and dy/dv, we need to express x and y in terms of v. We are given that u = x^2 + y^2 and v = 1/x + 1/y.

From v = 1/x + 1/y, we can rearrange it to express x and y in terms of v:

1/x = v - 1/y
x = 1 / (v - 1/y)

1/y = v - 1/x
y = 1 / (v - 1/x)

Now we can find dx/dv and dy/dv by differentiating x and y with respect to v:

dx/dv = d/dv (1 / (v - 1/y))
= 1 / (v - 1/y)^2 * (1/y^2)(dy/dv)
= 1 / (v - 1/y)^2 * (1/y^2) * (-1/y^2)
= -1 / [(v - 1/y) * y^4]

dy/dv = d/dv (1 / (v - 1/x))
= 1 / (v - 1/x)^2 * (1/x^2)(dx/dv)
= 1 / (v - 1/x)^2 * (1/x^2) * (-1/x^2)
= -1 / [(v - 1/x) * x^4]

Step 3: Use the chain rule to find dw/dv.

Now that we have dw/dx, dw/dy, dw/dz, dx/dv, and dy/dv, we can find dw/dv using the chain rule:

dw/dv = (dw/dx)(dx/dv) + (dw/dy)(dy/dv) + (dw/dz)(dz/dv)
= (-2) * (-1 / [(v - 1/y) * y^4]) + (-2) * (-1 / [(v - 1/x) * x^4]) + (1/z) * (dz/dv)

Since we are trying to show that dw/dv = 0, we can simplify this equation and set it equal to 0:

0 = (-2) * (-1 / [(v - 1/y) * y^4]) + (-2) * (-1 / [(v - 1/x) * x^4]) + (1/z) * (dz/dv)

Simplifying this equation further will involve substituting the expressions for x, y, and z in terms of v from the given variables.