# Thread: multivaribles 3??? caculus

1. ## multivaribles 3??? caculus

let f(x,y,z)=2x^2y^2z + 4xy^3z^2 find fxyz(2,1,1)

do i have to find partial dervtive of fx and fy and fz and sub the value in?

2. Originally Posted by zangestu888
let f(x,y,z)=2x^2y^2z + 4xy^3z^2 find fxyz(2,1,1)

do i have to find partial dervtive of fx and fy and fz and sub the value in?
Is this what you're trying to find?

$f_{xyz}(2, 1, 1)$?

3. I don't think so. You would have to find $f_x$ then partial derivate it with respect to $y$. Once you get this function, partial derivate with respect to $z$.
Then evaluate this function in $(2,1,1)$.
This method however doesn't work for all functions. For example the function $f(x,y)=xy \frac{x^2-y^2}{x^2+y^2}$.

4. In other words, you're finding

$\frac{\partial}{\partial z}\left(\frac{\partial}{\partial y}\left(\frac{\partial}{\partial x}\left(f(x, y, z)\right)\right)\right)$

and then evaluating it at $x = 2, y = 1, z = 1$.

5. Yeah the book says find
$
f_{xyz}(2, 1, 1)
$

does that mean i find fx then fxy then fxyz then find it at the point

6. Originally Posted by zangestu888
Yeah the book says find
$
f_{xyz}(2, 1, 1)
$

does that mean i find fx then fxy then fxyz then find it at the point
It certainly does.