With respect to y

• Sep 16th 2009, 09:42 PM
el123
With respect to y
$\displaystyle \frac{x^4}{1-xy}$

find derivative with respect to y.holding x constant.

$\displaystyle f_y=\frac{[(4x^3)(1-xy)]-[(x^4)(-x)]}{(1-xy)^2}$

$\displaystyle f_y= \frac{4x^3-4x^4y+x^5}{(1-xy)^2}$

Is that right or did i do that wrong?
Or is is suppose to be

$\displaystyle f_y=\frac{[(0)(1-xy)]-[(x^4)(-x)]}{(1-xy)^2}$

$\displaystyle f_y=\frac{x^5}{(1-xy)^2}$

???? Suggestions?
• Sep 16th 2009, 09:45 PM
Matt Westwood
The point here is "holding x constant".

So treat it as though it were:

$\displaystyle \frac d {dy} \left({\frac {c^4}{1-cy}}\right)$

or even:

$\displaystyle \frac d {dy} \left({\frac {a}{1-by}}\right)$

... which should be an elementary application of chain rule.
• Sep 16th 2009, 09:46 PM
seld
Quote:

Originally Posted by el123
$\displaystyle \frac{x^4}{1-xy}$

find derivative with respect to y.holding x constant.

$\displaystyle f_y=\frac{[(4x^3)(1-xy)]=[(x^4)(-x)]}{(1-xy)^2}$

Why is there an equal sign here?

Quote:

Originally Posted by el123
$\displaystyle f_y= \frac{4x^3-4x^4y+x^5}{(1-xy)^2}$

Is that right or did i do that wrong?
Or is is suppose to be

$\displaystyle f_y=\frac{[(0)(1-xy)]=[(x^4)(-x)]}{(1-xy)^2}$

$\displaystyle f_y=\frac{x^5}{(1-xy)^2}$

???? Suggestions?

Is that supposed to be the partial y for your very first expression?

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

because if you consider y as the only variable it's just power rule.

$\displaystyle x^4(1-xy)^{-1}$
• Sep 16th 2009, 09:47 PM
Matt Westwood
Quote:

Originally Posted by Matt Westwood
The point here is "holding x constant".

So treat it as though it were:

$\displaystyle \frac d {dy} \left({\frac {c^4}{1-cy}}\right)$

or even:

$\displaystyle \frac d {dy} \left({\frac {a}{1-by}}\right)$

... which should be an elementary application of chain rule.

In fact, in the last bit of your post I see you've done it right.
• Sep 16th 2009, 09:48 PM
el123
not meant to be equal , meant to be minus lol
• Sep 16th 2009, 09:48 PM
seld
Quote:

Originally Posted by Matt Westwood
In fact, in the last bit of your post I see you've done it right.

oh hmm you're right I didn't notice that either . . .

$\displaystyle f_y=\frac{x^5}{(1-xy)^2}$

That is right.
• Sep 16th 2009, 09:51 PM
el123
so i got it right in the second equation?
• Sep 16th 2009, 09:53 PM
Matt Westwood
Quote:

Originally Posted by el123
so i got it right in the second equation?

Yes - good job. Partial differentiation, it's more simple than you'd initially suspect.
• Sep 16th 2009, 09:54 PM
seld
if:
$\displaystyle f(x,y) = \frac{x^4}{1-xy}$

then:
$\displaystyle f_y=\frac{x^5}{(1-xy)^2}$

That solution for the partial y is correct.
• Sep 16th 2009, 09:58 PM
el123
gracias amigos!