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

i want to differentiate with respect to y.

So.

$\displaystyle [(0)(\ln(1-xy)]+[(3x^2)(-x)]$

which is $\displaystyle -3x^3$

Is that correct?

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- Sep 16th 2009, 10:20 PMel123another partial derivative
$\displaystyle 3x^2\ln(1-xy)$

i want to differentiate with respect to y.

So.

$\displaystyle [(0)(\ln(1-xy)]+[(3x^2)(-x)]$

which is $\displaystyle -3x^3$

Is that correct? - Sep 16th 2009, 10:49 PMseld
again you're running into this same problem where you want to do a partial derivative in terms of y, but you're not holding x as a constant.

what type of function of y are you needing to differentiate (ie. polynomial, quotient etc.) - Sep 16th 2009, 11:03 PMredsoxfan325
Just treat $\displaystyle x$ like a number. $\displaystyle 3x^2$ and $\displaystyle x$ are just constants.

If you were derivating $\displaystyle c_1\ln(1-c_2 y)$, what would you get? You'd get $\displaystyle \frac{c_1}{1-c_2 y}\cdot -c_2$. Just extend that notion to this problem. - Sep 17th 2009, 12:24 AMel123
so its $\displaystyle \frac{3x^3}{1-xy}$

Is that what your saying? - Sep 17th 2009, 06:15 AMredsoxfan325
You have to multiply that by $\displaystyle \frac{d}{dx}[1-xy]=-x$, to complete the chain rule, so the answer is $\displaystyle \frac{-3x^3}{1-xy}$.