1. ## Partial Derivative

Hey, I am really rusty on my multivariable calc stuff and cannot seem to get the right answer to...

$\displaystyle f(x,y,z)=\tan^{-1}(\frac{1}{xy^{2}z^{3}})$ with respect to x and with respect to y.

Help? I don't know what techniques to imply. I just went ahead and did it as if it were just a single variable function since I only needed it with respect to x and also with respect to y.

Thanks

2. If you will show your work here..then it will be easier to tell

3. $\displaystyle \frac{\partial f}{\partial x} = \frac {1}{1+(\frac{1}{xy^{2}z^{3}})^{2}}* \frac {\partial}{\partial x} \frac{1}{ xy^{2}z^{3}}$

$\displaystyle = \frac {1}{1+\frac{1}{x^{2}y^{4}z^{6}}}* \frac{\text{-}1}{ x^{2}y^{2}z^{3}}$

$\displaystyle = \frac {-1}{x^{2}y^{2}z^{3}+ \frac{1}{y^{2}z^{3}}}$

$\displaystyle = \frac {-y^{2}z^{3}}{x^{2}y^{4}z^{6} + 1}$

4. $\displaystyle f(x,y,z)=\tan^{-1}(\frac{1}{xy^{2}z^{3}})$

You want $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$

$\displaystyle u = \frac{1}{xy^{2}z^{3}}$

$\displaystyle \frac{\partial}{\partial u}(\frac{1}{tan{(u)}})$

Then when when calculating $\displaystyle u'$ remember that

$\displaystyle \frac{\partial}{\partial x}(xy^{2}z^{3})=1*y^{2}z^{3}$

and that

$\displaystyle \frac{\partial}{\partial y}(xy^{2}z^{3})=2xyz^{3}$

As long as this isn't an implicit function, you treat the other variables as constants while deriving for one.

Also remember the derivatives rules for fractions

$\displaystyle (\frac{f}{g})'=\frac{f'*g-f*g'}{g^{2}}$

And the derivative of tangent

$\displaystyle tan(u)' = u'*sec^{2}(u)$

That's all you need to know.