# First and second order partial derivatives

Hello,

I was wondering if I could get some help with a question I have.

## Homework Statement

We are asked to find the first and second order partial derivatives of

f(x,y) = x^2 - y^2 - 4x^2/(y - 1)^2 (sorry, I dont know how to write this in latex).

I am not really sure how to get started with this, so any help at all would be fantastic.

Thanks

Sean Feely

Related Calculus and Beyond Homework Help News on Phys.org
Just calculate the derivatives in respect to one variable while treating the other variable as if it was a constant. Do this for both variables in different orders and you should get 6 derivatives (2 of the 1st order and 4 of the 2nd order) although two of them will be the same.

djeitnstine
Gold Member
We are asked to find the first and second order partial derivatives of

f(x,y) = x^2 - y^2 - 4x^2/(y - 1)^2 (sorry, I dont know how to write this in latex).
First and second order partial derivatives with respect to which variable? Both?

Pengwuino
Gold Member
You are being asked to find $$\frac{{\partial f(x,y)}}{{\partial x}},\frac{{\partial f(x,y)}}{y},\frac{{\partial ^2 f(x,y)}}{{\partial x^2 }},\frac{{\partial ^2 f(x,y)}}{{\partial y^2 }}$$ and possibly the mixed partial derivative, $$\frac{{\partial f(x,y)}}{{\partial x\partial y}}$$.

Let's say you have a function $$f(x,y) = 3x + 2y + 6xy$$. To find the first partial with respect to x, that is $$\frac{{\partial f(x,y)}}{{\partial x}}$$, you simply pretend y is a constant. Thus, you find the partial with respect to x is $$\frac{{\partial f(x,y)}}{{\partial x}} = 3 + 6y$$. Since y is constant, 2y is a constant with respect to x, and thus it goes away when you differentiate with respect to x. The second derivative from there is obviously 0 since the derivative of 3 is 0 and the derivative of 3y is 0 because y is a constant just like the number 3. When you take partials with respect to y, you do the same thing except with x as constant. The first derivative with respect to y would be 2 + 6x and again, the second derivative would be 0.

The mixed partial is a bit different as you switch the order. Either way is fine but you can first take the derivative with respect to x which gives you 3 + 6y. Then you have to take the derivative of y, and since y is no longer a constant, you see that the mixed partial is simply equal to 6. It actually does not matter which way you take your mixed partials, x first y second, y first x second in most cases.