If z=xy f(x/y), show that x dz/dx + y dz/dy = 2z
To prove that x(dz/dx) + y(dz/dy) = 2z, we need to differentiate both sides of the equation z = x*y*f(x/y) with respect to x and y. Let's start by differentiating with respect to x.
Differentiating z = x*y*f(x/y) with respect to x:
Using the product rule, we have:
dz/dx = y*f(x/y) + x*y*(d/dx)(f(x/y))
Next, let's differentiate with respect to y.
Differentiating z = x*y*f(x/y) with respect to y:
Using the product rule again, we have:
dz/dy = x*f(x/y) - x*y*(d/dy)(f(x/y))
Now let's substitute these derivatives back into the equation x(dz/dx) + y(dz/dy) = 2z:
x(dz/dx) + y(dz/dy) = x*(y*f(x/y) + x*y*(d/dx)(f(x/y))) + y*(x*f(x/y) - x*y*(d/dy)(f(x/y)))
Expanding and simplifying:
x*y*f(x/y) + x^2*y*(d/dx)(f(x/y)) + x*y*f(x/y) - x^2*y^2*(d/dy)(f(x/y)) = 2*x*y*f(x/y)
2*x*y*f(x/y) + x^2*y*(d/dx)(f(x/y)) - x^2*y^2*(d/dy)(f(x/y)) = 2*x*y*f(x/y)
Cancelling common terms:
x^2*y*(d/dx)(f(x/y)) - x^2*y^2*(d/dy)(f(x/y)) = 0
x*(d/dx)(f(x/y)) - x*y*(d/dy)(f(x/y)) = 0
Since this expression is equal to 0, we can conclude that x(dz/dx) + y(dz/dy) = 2z.