Total Differential

Find the total differential:

$\displaystyle u=arcsen \sqrt\frac{xz}{y}$
$\displaystyle du=?$

My teacher give to us this formule: $\displaystyle \Delta f (x,y) \approx \frac{\delta f}{\delta x} |p\Delta x + \frac{\delta f}{\delta y}|p\Delta y + \frac{\delta f}{\delta z}|p\Delta z$

but in this exercice he just dont give the point "P" and the valous of $\displaystyle \Delta n$
i use this formule: $\displaystyle \Delta f (x,y) \approx \frac{\delta f}{\delta x} + \frac{\delta f}{\delta y} + \frac{\delta f}{\delta z}$
and the result was:
part 1: http://i.imgur.com/W46hP.jpg
part 2: http://i.imgur.com/uYTek.jpg
is this correct??

Hi again Chipset3600! :)

The proper formula is:

$\displaystyle df(x,z,y) = {\partial f \over \partial x}dx + {\partial f \over \partial y}dy + {\partial f \over \partial z}dz$

I have some trouble understanding your scans, but they do not look right.

Let's define $\displaystyle g(x,y,z)=\sqrt{xz \over y}$.
And let's define $\displaystyle u(v)=\arcsin v$.

What is $\displaystyle {\partial g \over \partial x}$?

And $\displaystyle {du \over dv}$?

dg/dx is the first derivative of G as a function of x

and du/dv is the total diferencial of function u, rite??

Quote:

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Originally Posted by Chipset3600

dg/dx is the first derivative of G as a function of x

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Yes.
It's called the partial derivative of g as a function of x.

Quote:

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Originally Posted by Chipset3600

and du/dv is the total diferencial of function u, rite??

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It's just the derivative of u with respect to v.

well did a mistake in the formule, buts this means that my result isn's correct ? :(

Well, I couldn't make sense of your first scan, so just now I skipped to the second one.

And yes! :)
Your partial derivatives are correct (except for $\displaystyle \partial u \over \partial y$ that should have a minus sign).

However, for the total derivative du you shouldn't have added them as you did.
Each of them should be multiplied by dx, dy, and dz respectively.

One more time thank you very match!
We most talk!