Math Is Fun Forum

mike

Guest

Partial diffrentiation

this is one topic that i drag at.
it is Partial diffrentiation

i have this question

ƒ(x,y) = e^2xy^2 cos(3y)

find ¶f/¶x   and   ¶f/¶y

irspow

Member

Registered: 2005-11-24

Posts: 1,055

Re: Partial diffrentiation

I am no good either, however taking partial derivatives isn't that bad.  Just treat the variable that you are not differentiating with respect to as a constant.

  For example:

  f(x,y) = 2xy + x² - y²

  ∂f/∂y = 2x - 2y

  ∂f/∂x = 2y + 2x

  I hope that helps, because the problem you have above, if typed correctly, is a little ugly.

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Ricky

Moderator

Registered: 2005-12-04

Posts: 3,791

Re: Partial diffrentiation

ƒ(x,y) = e^2xy^2 cos(3y)

You need to use parenthesis.  is it e^(2), e^(2x), etc...

Assuming it's e^(2x) * y^2 * cos(3y):

¶f/¶x:

Assume that y is a constant.  So y^2*cos(3y) is a constant.  Let y^2*cos(3y) = k.  The equation then becomes:

k*e^(2x)

Taking the derivative, this is: 2ke^(2x), which is 2(y^2*cos(3y))*e^(2x).

Now try the same for y.

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"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

kubai paul

Guest

Re: Partial diffrentiation

:Dhelp me to differentiate the following

y=x^lnx find dy/dx

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: Partial diffrentiation

Replace x by e^(ln x). Then you can use power laws to write it as e^(f(x)), for some f, and differentiate that without too much trouble.

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Why did the vector cross the road?
It wanted to be normal.

LampShade

Member

Registered: 2009-02-22

Posts: 23

Re: Partial diffrentiation

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