Partial differentation

**anderson** · March 6th 2010, 11:12 AM

Hi everyone
Need help with this question,lifesaver guaranteed.

Find dz/dx & dz/dy for
4ln(4xyz)+sin(xz2)=0
This is the working but i'm not sure if it is corect,need help.
$F_x=4*\frac {4yz}{4xyz}+z^2cosxz^2=\frac{4}{x}+z^2cosxz^2$

$F_z=4*\frac{4xy}{4xyz}+2zxcosxz^2=\frac{4}{z}+2zxc osxz^2$
$F_y=4*4xz$

Thank you for all your help & support.

**drumist** · March 6th 2010, 12:26 PM

So, our starting equation is

$4 \ln (4xyz) + \sin (xz^2) = 0$

When computing $\frac{\partial z}{\partial x}$ , we are considering z to be a function of x, i.e., z(x). So, any time we have a term containing z multiplied with a term containing x, we must use the product rule.

So now we differentiate:

$\frac{\partial}{\partial x}\left[4 \ln (4xyz) + \sin (xz^2)\right] = \frac{\partial}{\partial x}(0)$

$4 \frac{\frac{\partial}{\partial x}(4xyz)}{4xyz} + \cos (xz^2) \frac{\partial}{\partial x} (xz^2) = 0$

Looking at these two partials:

$\frac{\partial}{\partial x}(4xyz) = 4y\frac{\partial}{\partial x}(xz) = 4y\left(x \frac{\partial z}{\partial x} + z\right)$

and

$\frac{\partial}{\partial x}(xz^2) = 2xz \frac{\partial z}{\partial x} + z^2$

Thus:

$4 \frac{4y(x \frac{\partial z}{\partial x} + z)}{4xyz} + \cos (xz^2) \left(2xz \frac{\partial z}{\partial x} + z^2\right) = 0$

At this stage, you can simplify and solve for $\frac{\partial z}{\partial x}$ .

I'll leave the other partial for you to solve using the same technique.

**HallsofIvy** · March 6th 2010, 02:14 PM

Did you notice that ln(xyz)= ln(x)+ ln(y)+ ln(z)?

That's why $\frac{\partial ln(xyz)}{\partial x}= \frac{yz}{xyz}= \frac{1}{x}= \frac{\partial ln(x)}{\partial x}$

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