Total Differential

**Chipset3600** · December 18th 2011, 11:43 AM

Find the total differential:

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

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

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

**ILikeSerena** · December 18th 2011, 12:04 PM

Hi again Chipset3600!

The proper formula is:

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

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

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

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

And ${du \over dv}$ ?

**Chipset3600** · December 18th 2011, 12:11 PM

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

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

**ILikeSerena** · December 18th 2011, 12:16 PM

Originally Posted by Chipset3600

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

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

Originally Posted by Chipset3600

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

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

**Chipset3600** · December 18th 2011, 12:18 PM

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

**ILikeSerena** · December 18th 2011, 12:30 PM

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 $\partial u \over \partial y$ 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.

**Chipset3600** · December 18th 2011, 12:35 PM

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

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