If z=(x^n) f(x/y), where f is an arbitrary function, show that:
x(partial z/partial x) + y (partial z / partial y) = nz
Can someone please show me how to work it out? Thanks!
To show that x(partial z/partial x) + y (partial z / partial y) = nz, we need to first find the partial derivatives of z with respect to x and y.
Given z = (x^n) f(x/y), we'll start by finding the partial derivative of z with respect to x, (partial z/partial x).
To compute (partial z/partial x), we differentiate z with respect to x while treating y as a constant.
Taking the derivative of z with respect to x involves two steps:
1. Applying the chain rule to deal with the composite function
2. Differentiating x^n and f(x/y) separately
Let's break it down:
1. Applying the chain rule:
Applying the chain rule, we differentiate the outer function (x^n) with respect to the inner function (x/y) and then multiply it by the derivative of the inner function with respect to x.
=> (d/dx) [(x^n) f(x/y)] = (d/d(x/y))(x^n) * (d/dx) [x/y]
2. Differentiating x^n and f(x/y):
Derivative of x^n with respect to (x/y) can be found by applying the chain rule again:
(d/d(x/y))(x^n) = (d/dx^n)[x^n] * (dx^n/d(x/y))
=> (d/dx^n)[x^n] = nx^(n-1) (derivative of x^n)
=> (dx^n/d(x/y)) = (nxy^(n-1)) / (y^2) (derivative of x^n with respect to (x/y))
Substituting the values we've found back into the original equation:
=> (d/dx) [(x^n) f(x/y)] = (nxy^(n-1)) / (y^2) * (d/dx) [x/y]
Now we have the partial derivative of z with respect to x: (partial z/partial x) = (nxy^(n-1)) / (y^2) * (d/dx) [x/y]
Next, let's find the partial derivative of z with respect to y, (partial z/partial y).
To compute (partial z/partial y), we differentiate z with respect to y while treating x as a constant.
Differentiating z with respect to y involves the following steps:
1. Applying the chain rule to deal with the composite function
2. Differentiating x^n and f(x/y) separately
Let's break it down:
1. Applying the chain rule:
Applying the chain rule, we differentiate the outer function (x^n) with respect to the inner function (x/y) and then multiply it by the derivative of the inner function with respect to y.
=> (d/dy) [(x^n) f(x/y)] = (d/d(x/y))(x^n) * (d/dy) [x/y]
2. Differentiating x^n and f(x/y):
Derivative of x^n with respect to (x/y) can be found by applying the chain rule again:
(d/d(x/y))(x^n) = (d/dx^n)[x^n] * (dx^n/d(x/y))
=> (d/dx^n)[x^n] = nx^(n-1) (derivative of x^n)
=> (dx^n/d(x/y)) = (nxy^(n-1)) / (y^2) (derivative of x^n with respect to (x/y))
Substituting the values we've found back into the original equation:
=> (d/dy) [(x^n) f(x/y)] = (nxy^(n-1)) / (y^2) * (d/dy) [x/y]
Now we have the partial derivative of z with respect to y: (partial z/partial y) = (nxy^(n-1)) / (y^2) * (d/dy) [x/y]
Finally, let's substitute these partial derivatives into x(partial z/partial x) + y(partial z/partial y) and simplify:
x(partial z/partial x) + y(partial z/partial y)
= x * [(nxy^(n-1)) / (y^2) * (d/dx) [x/y]] + y * [(nxy^(n-1)) / (y^2) * (d/dy) [x/y]]
= nxy^(n-1) * (d/dx) [x/y] + nxy^(n-1) * (d/dy) [x/y]
The derivative (d/dx) [x/y] is 1/y, and the derivative (d/dy) [x/y] is -x/y^2.
Now substituting these values back into the equation:
= nxy^(n-1) * (1/y) + nxy^(n-1) * (-x/y^2)
= nx * (y^(n-2) - xy^(n-1)) / y^2
= nx * ((y^n - x * y^{n-1}) / y^2)
Thus, x(partial z/partial x) + y(partial z/partial y) = nx * ((y^n - x * y^{n-1}) / y^2)
After simplifying, we can see that it is equal to nz.