partial derivative

**el123** · September 15th 2009, 11:45 PM

I have

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

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

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

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

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

**seld** · September 15th 2009, 11:51 PM

There's only 1 function in terms of y.

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

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

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

The only variable is y, does that help?

**el123** · September 16th 2009, 12:04 AM

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

??

**seld** · September 16th 2009, 12:07 AM

err you dropped some constants.

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

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

subsitute x back in for a:

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

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

**el123** · September 16th 2009, 12:28 AM

so if i had $x^3ln(1-xy)$

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

Would that be correct?

**seld** · September 16th 2009, 12:34 AM

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

$(x^3*\frac{1}{1-xy})*\frac{\partial}{\partial y}<br /> (1-xy)$

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

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

Does that make sense to you?

**el123** · September 16th 2009, 12:53 AM

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

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