1. ## partial derivative

I have

$\displaystyle x^2\ln(xy+4)$

$\displaystyle f_x = 2x\ln(xy+4)+\frac{x^2y}{xy+4}$

Thats correct i think but this next one im confused about.

$\displaystyle f_y = 2x\ln(xy+4)+\frac{xy^2}{xy+4}$

Is that correct , if not can someone explain to me how to differentiate that function with respect to y?

2. There's only 1 function in terms of y.

$\displaystyle x^2\ln(xy+4)$

if you view x as a constant let's replace x with a.

$\displaystyle a^2\ln(ay+4)$

The only variable is y, does that help?

3. so $\displaystyle f_y = \frac{1}{y+4}$

??

4. err you dropped some constants.

$\displaystyle f(x)=a^2\ln(ay+4)$

$\displaystyle f_y= a^2 \frac{1}{ay+4}*a$

subsitute x back in for a:

$\displaystyle f_y=x^2\frac{1}{xy+4}*x$

$\displaystyle =x^3\frac{1}{xy+4}$

5. so if i had $\displaystyle x^3ln(1-xy)$

$\displaystyle f_y =\frac{x^4}{1-xy}$

Would that be correct?

6. um . . . chain rule the constant in front of y is -x

$\displaystyle (x^3*\frac{1}{1-xy})*\frac{\partial}{\partial y} (1-xy)$

$\displaystyle (\frac{x^3}{1-xy})*(-x)$

$\displaystyle \frac{-x^4}{1-xy}$

Does that make sense to you?

7. yeah woops forgot the negative , yip that makes sense , cheers mate.