Let z = f(x,y) be a homogeneous function in x and y of degree *n, *and having continuous partial derivatives, then

x(∂z/∂x+ y(∂z/∂z) = *n z*

**Proof:**

Since z is a homogeneous function of degree *n, *we can write

F(x,y) = x^{n} f(y/x)

Substituting (y/x) by t, we get, f(x,y) = x^{n} f(t)

Differentiating both the sides partially with respect to x and y respectively, we have,

= ∂z/∂x= nx^{n-1}f(t) + x^{n} f ’ (t) (∂t/∂x)

= nx^{n-1}f(y/x) + x^{n} f’ (y/x) (-y/x²) (∴ t =y/x)

= nx^{n-1}f(y/x) – x^{n} y x ^{n-2 }, f’ (y/x)

Again, we have, ∂t/∂y= xn f ’ (t) (1/x)

= nx^{n-1}f ’ (y/x)

Multiplying ∂z/∂x by x and ∂z/∂y by *y *and then adding their products we get,

x(∂z/∂x) +y(∂z/∂y) = n x^{n} f(y/x) = n *f*(x,y) =nz

Hence, x(∂z/∂x) + y(∂z/∂y) = nz Proved.