# first partial derivatives

• February 17th 2009, 02:00 PM
viet
first partial derivatives
Quote:

Find the first partial derivatives of $f(x,y) = \frac{2x-4y}{2x+4y}$ at the point (x,y) = (2, 2).

\frac{\partial f}{\partial x}(2,2) =__
\frac{\partial f}{\partial y}(2,2)=__
I need help solving this problem.
• February 17th 2009, 02:18 PM
Mush
Quote:

Originally Posted by viet
I need help solving this problem.

When you take a partial derivatives with respect to a variable, you treat all other variables as constants.

$f(x,y) = \frac{2x - 4y}{2x+4y}$

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \bigg(\frac{2x - 4y}{2x+4y}\bigg)$

Use the quotient rule:

$= \frac{(2x+4y)\frac{\partial}{\partial x}(2x-4y) - (2x-4y) \frac{\partial}{\partial x}(2x+4y)}{(2x+4y)^2}$

$= \frac{(2x+4y)(2) - (2x-4y)(2)}{(2x+4y)^2}$

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \bigg(\frac{2x - 4y}{2x+4y}\bigg)$

Again use the quotient rule:

$= \frac{(2x+4y)\frac{\partial}{\partial y}(2x-4y) - (2x-4y) \frac{\partial}{\partial y}(2x+4y)}{(2x+4y)^2}$

$= \frac{(2x+4y)(-4) - (2x-4y)(4)}{(2x+4y)^2}$

Now plug in the points given in the question.