$\displaystyle 3x^2 \ln(1-xy)$

the partial derivatvie with respect to y is

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

is that right?

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- Sep 18th 2009, 06:30 PMel123Partial 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 PMProve It
- Sep 18th 2009, 06:39 PMel123
but i thought you treat x as a constant so its derivative is 0?

- Sep 18th 2009, 06:44 PMProve It
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 PMel123
true to that , thank you very much !