# Partial derivative

• Sep 18th 2009, 06:30 PM
el123
Partial derivative
$\displaystyle 3x^2 \ln(1-xy)$

the partial derivatvie with respect to y is

$\displaystyle \frac{x}{1-xy}$

is that right?
• Sep 18th 2009, 06:35 PM
Prove It
Quote:

Originally Posted by el123
$\displaystyle 3x^2 \ln(1-xy)$

the partial derivatvie with respect to y is

$\displaystyle \frac{x}{1-xy}$

is that right?

Not quite.

It's actually

$\displaystyle 3x^2\left(\frac{-x}{1 - xy}\right)$

$\displaystyle = -\frac{3x^3}{1 - xy}$.
• Sep 18th 2009, 06:39 PM
el123
but i thought you treat x as a constant so its derivative is 0?
• Sep 18th 2009, 06:44 PM
Prove It
Quote:

Originally Posted by el123
but i thought you treat x as a constant so its derivative is 0?

Yes, it is treated like a constant.

What do you do to differentiate a constant times a function?

It's the constant times the derivative of that function.

i.e. $\displaystyle \frac{d}{dx}\left[c\cdot f(x)\right] = c\cdot f'(x)$.

So in your case, you have $\displaystyle 3x^2 \ln{(1 - xy)}$

The $\displaystyle 3x^2$ is your constant.

So the partial derivative will be

$\displaystyle 3x^2 \frac{\partial}{\partial y}\left[\ln{(1 - xy)}\right]$.
• Sep 18th 2009, 06:46 PM
el123
true to that , thank you very much !