Question

Find du/dy where u=wx/yz

Answer 1

\[u=\frac{wx} {yz} \]

Answer 2

PLZ HELP ME.

Answer 3

I'm starting to think\[u=wxy^{-1}z^{-1}\\\frac{du}{dy}=wxz^{-1}\times1\times y^{-1-1}\\\frac{du}{dy}=\frac{wx}{y^2z}\]Could this be right?

Answer 4

please note that we have:
\[\frac{d}{{dy}}\left( {\frac{1}{y}} \right) = - \frac{1}{{{y^2}}}\]

Answer 5

or:
\[\frac{d}{{dy}}\left( {\frac{1}{y}} \right) = \frac{d}{{dy}}\left( {{y^{ - 1}}} \right) = - 1 \cdot {y^{ - 1 - 1}} = - {y^{ - 2}}\]

Answer 6

ah HA! Yes, that's what I missed. Thanks @Michele_Laino!