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First and second order partial derivatives

  • Thread starter feely
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
180
4
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.
 
  • #3
djeitnstine
Gold Member
614
0
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?
 
  • #4
Pengwuino
Gold Member
4,989
15
You are being asked to find [tex]\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 }}[/tex] and possibly the mixed partial derivative, [tex]\frac{{\partial f(x,y)}}{{\partial x\partial y}}[/tex].

Let's say you have a function [tex]f(x,y) = 3x + 2y + 6xy[/tex]. To find the first partial with respect to x, that is [tex]\frac{{\partial f(x,y)}}{{\partial x}}[/tex], you simply pretend y is a constant. Thus, you find the partial with respect to x is [tex]\frac{{\partial f(x,y)}}{{\partial x}} = 3 + 6y[/tex]. 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.
 

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