# Math Help - partial derivatives

1. ## partial derivative

show that fxy =fyx for f(x,y)= x sqrt(y-x^2)

when i find do this they are not equal but from the definition they are supposed to be...can some one show me the work to compare it to mine so i can see where i messed up???

2. Originally Posted by chris25
show that fxy =fyx for f(x,y)= x sqrt(y-x^2)

when i find do this they are not equal but from the definition they are supposed to be...can some one show me the work to compare it to mine so i can see where i messed up???
$f(x, y) = x \sqrt{y - x^2}$

$f_x = \sqrt{y - x^2} + x \cdot \frac{1}{2 \sqrt{y - x^2}} \cdot -2x$

$f_x = \sqrt{y - x^2} - \frac{x^2}{\sqrt{y - x^2}}$

$f_{xy} = \frac{1}{2\sqrt{y - x^2}} + \frac{x^2}{2(y - x^2)^{3/2}}$

You do $f_{yx}$.

-Dan

3. ## partial derivatives

for f(x,y) = x sqrt (y-x^2)
find fy and fyx

i think im messing up fy which is messing up fyx can anyone help me differentiate these???

4. Originally Posted by chris25
for f(x,y) = x sqrt (y-x^2)
find fy and fyx

i think im messing up fy which is messing up fyx can anyone help me differentiate these???
how did you do it?

5. used chain rule and received x/2sqrty-x^2
after that i go wrong

6. Originally Posted by chris25
used chain rule and received x/2sqrty-x^2
after that i go wrong
That is correct for $f_y$. So now do your x derivative. Use the quotient rule:
$f_{yx} = \frac{(x)' \cdot (2\sqrt{y - x^2)} - (x) \cdot (2 \sqrt{y - x^2})'}{(2 \sqrt{y - x^2})^2}$
where the primes represent derivatives with respect to x.

-Dan

7. Originally Posted by chris25
show that fxy =fyx for f(x,y)= x sqrt(y-x^2)

when i find do this they are not equal but from the definition they are supposed to be...
Is it always supposed to be? Consider $D$ where $D\equiv\left\{(x,y):y