# Thread: With respect to y

1. ## With respect to y

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

find derivative with respect to y.holding x constant.

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

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

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

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

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

???? Suggestions?

2. The point here is "holding x constant".

So treat it as though it were:

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

or even:

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

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

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

find derivative with respect to y.holding x constant.

$f_y=\frac{[(4x^3)(1-xy)]=[(x^4)(-x)]}{(1-xy)^2}$
Why is there an equal sign here?

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

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

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

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

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

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

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

$x^4(1-xy)^{-1}$

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

So treat it as though it were:

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

or even:

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

... 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.

5. not meant to be equal , meant to be minus lol

6. 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 . . .

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

That is right.

7. so i got it right in the second equation?

8. 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.

9. if:
$
f(x,y) = \frac{x^4}{1-xy}
$

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

That solution for the partial y is correct.

10. gracias amigos!