# Partial derivatives

• Mar 6th 2007, 05:34 PM
clockingly
Partial derivatives
I have to find the partial derivative for:

F(x,y) = sin(xy) at (1, pi) with respect to x and with respect to y

I'm really not getting this concept of partial derivatives, so if someone could provide me with an example of how to calculate one, I would appreciate it!
• Mar 6th 2007, 05:58 PM
Jhevon
Quote:

Originally Posted by clockingly
I have to find the partial derivative for:

F(x,y) = sin(xy) at (1, pi) with respect to x and with respect to y

I'm really not getting this concept of partial derivatives, so if someone could provide me with an example of how to calculate one, I would appreciate it!

the concept of partial derivatives goes like this. when you have a multivariable function and you want to find the partial derivative with respect to one of those variables, you treat all the other variables as constants and find the derivative of the desired variable normally.

F(x,y) = sin(xy)

dF/dy = xcos(xy) ........i do the chain rule treating x as a number. if x is a number, the derivative of xy is x, so i multiply by x to satisfy the chain rule

dF/dx = ycos(xy)

note, these are partial derivatives, so the d's should be written like lower case delta's
• Mar 6th 2007, 06:03 PM
clockingly
Thanks!

What if you have something like an e raised to a power, though? What happens then?

I have another problem...finding the partial derivative of the following w/ respect to x and y...

F(x,y) = 5e^(2pi/y)
• Mar 6th 2007, 06:23 PM
Jhevon
Quote:

Originally Posted by clockingly
Thanks!

What if you have something like an e raised to a power, though? What happens then?

I have another problem...finding the partial derivative of the following w/ respect to x and y...

F(x,y) = 5e^(2pi/y)

like i said, when you take the derivative with respect to x, you treat y as a constant. when you take the derivative with respect to y, you treat x as a constant.

F(x,y) = 5e^(2pi/y)

Taking the derivative with respect to x. Notice there is no x in the function. therefore, if we treat y as a constant, this function is just a constant. and so...

dF/dx = 0

taking derivative with respect to y

dF/dy = 5*(-2pi/y^2)e^(2pi/y) = [-10pi*e^(2pi/y)]/y^2

i used the chain rule here, do you understand what i did?
• Mar 6th 2007, 07:01 PM
clockingly
Sorry...I don't really understand where the y^2 comes from...

dF/dy = 5*(-2pi/y^2)e^(2pi/y) = [-10pi*e^(2pi/y)]/y^2
• Mar 6th 2007, 07:10 PM
Jhevon
Quote:

Originally Posted by clockingly
Sorry...I don't really understand where the y^2 comes from...

dF/dy = 5*(-2pi/y^2)e^(2pi/y) = [-10pi*e^(2pi/y)]/y^2

remember, for the chain rule, you multiply by the derivative of what e is raised to. e is raised to 2pi/y = 2pi*y^-1. the derivative of this is -2pi*y^-2 or -2pi/y^2
• Mar 6th 2007, 07:16 PM
clockingly
Oh! I see.

Okay, so if I had the equation F(x,y) = x^2 + 3x + y^2 - 4y + 7,

For respect to x it would be: 2x + 3

and

For respect to y it would be: 2y - 4

if I'm not mistaken?
• Mar 6th 2007, 07:22 PM
AfterShock
Quote:

Originally Posted by clockingly
Oh! I see.

Okay, so if I had the equation F(x,y) = x^2 + 3x + y^2 - 4y + 7,

For respect to x it would be: 2x + 3

and

For respect to y it would be: 2y - 4

if I'm not mistaken?

Correct.