If z= sqrt[(x^2)y + x(y^2)] and (x, y) changes from (5, 20) to (5.2, 20.4), compare the values of delta z (triangle z) and dz.

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If z= sqrt[(x^2)y + x(y^2)] and (x, y) changes from (5, 20) to (5.2, 20.4), compare the values of delta z (triangle z) and dz.

Mathematics
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Okay, I'll take a quick crack at it
kk
First off, what is this question for?

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Other answers:

What kind of class?
calculus 2
Oohhhh... F... Okay, I'm glad I asked. I'm only in Trig. Coincidentally, my teacher was saying how important trig was for calc 2, but I'm skeptical I can be helpful.
Considering how dead it is here, I doubt anyone else is on btw...
I actually came for help on my own homework, but noone responded :P
thas cool ill get it later
Hmm... I've my brother, who took calc 2 is up, actually. Lemme Harass er ...ask him.
What does d in dz mean, btw?
yo nvm. ill post it again later i think i got it. thnxs
Alrighty. Take it easy. I'm gonna try to finish my homework now. <_< ^^
if some1 can solve it that would b good to check my answer thnks
I'm working on it
ok thanks
What did you get for dz?
i stopped working on it i moved onto the next question
well you know x=5 y=20 dx=.2 and dy=.4 differentiate z to get dz and then you can plug in the values of x,y,dx,dy resulting in dz
can u show me the full solution?
do you know what the correct answer is? just want to check it with mine before I load the full solution
no i actually dont
\[dz=[1/2(x^2y+xy^2)^{-1/2}(2xy+y^2)]dx\] \[+ [1/2(x^2y+xy^2)^{-1/2}(x^2+2xy)]\] \[x=5, dx=.2, y=20, dy=.4\] Plug these values in and let me know what you get
forgot the dy, sorry \[[1/2(x^2y+xy^2)−1/2(x^2+2xy)]dy\]
gotdamit i messed it up again hold on
\[+[1/2(x^2y+xy^2)^{−1/2}(x^2+2xy)]dy\]
Here is the entire equation with any typos, sorry for the confusion \[dz=[1/2(x^2y+xy^2)^{−1/2}(2xy+y^2)]dx\] \[+[1/2(x^2y+xy^2)^{−1/2}(x^2+2xy)]dy\] \[x=5,dx=.2,y=20,dy=.4\]

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