Partial derivative...?

Question

I'm confused with what happens when you have to find the partial derivative of something like f(x,y)=2x^2 - 5xy with respect to both x and y, what exactly do you do with the term -5xy since it has both variables in it?

Yes, but what happens to the xy in the 5xy term? Why does it dissapear? please be very specific, thanks..

Answer 1

You hold one variable constant so if you are taking it with respect to x then the derivative would be 4x-5. The other way around it would be -5.

Answer 2

deriviative is 4x-(5xy'+5y+xy)
use product rule for 5xy, use x's actual derivative and write derivative of y as y' and leave it alone after that.
if you didnt know product rule is derivitive of one term times the other plus the derivitive of the other term times the first term thats hard to understand so der. of xy is x'y+xy' which is normally written as y+xy' in terms of x since derivative of x is 1

You can also use the chain rule after pulling an x out so that it is
x(2x-5y) wich then you would use the product rule anyway and get (2x-5y)+x(2+5y') which is 4x -5y+2x+5y' which with a little more work you can show equals the derivative of the first example i showed you.

Answer 3

Differentiating wrt x, you treat every y as a constant. So 5y is a constant so 5yx is no different from say 23x, its derivative is 5y and the derivative of your function is 4x-5y. Now when you differentiate wrt y, then x becomes a constant, so is 2x^2 and 5x and so the derivative is -5x.

Answer 4

when we get the partial derative of a fuunction (of two varibles or more) ....we differentiate to the varibale we want and concider other variables as constants


For the given example
F(x,y) = 2 x² - 5xy

to get ∂ F(x,y) / ∂x we differentiant to the variable x considering y as a constant
∂ F(x,y) / ∂x = 4x - 5y

to get ∂ F(x,y) / ∂y we differentiant to the variable yconsidering x as as constant
∂ F(x,y) / ∂y = 2x² - 5x

Answer 5

I did that like 4 years ago, may not be right. You first derivative x and y would be equal y' (y prime) then derivative the y term and sub it in the y' in the first derivative.

Answer 6

When differentiating wrt x treat all y's as a constant and when differentiating wrt y treat all x's as a constant.

In general when you are doing a partial derivative wrt a variable, say x, you treat all other variables as constants.

Answer 7

given f(x,y)=2x^2-5xy
differentite w r t x we get,
4x-5(x.0+y>1) here we first treat x as constant and differentiate y it becomes zero,y as constant and differentiate x it becomes 1
answer is 4x-5y

differentiate w r t y we get
0-5(x.1+y.0) here we first treat x as constant and differentiate y it becomes 1 y as constant and differentiate x it becomes 0

answer is -5x
we apply the product formula d/dx(uv)=u.dv+v.du

Answer 8

when you do a full derivative you shall have dy/dx not zero. That is
df/dx = 4x-5(d/dx(xy))
= 4x - 5(ydx/dx+xdy/dx)
= 4x-5(y+xdy/dx)
but when you have partial derivative wrt x
you have y constant so
df/dx(partial derivaltive) = 4x-5y
df/dy(patrial drivative) = - 5x as x is constant

Basically tou treat all other variables as constants