Paul's Online Notes
Home / Calculus III / Partial Derivatives / Partial Derivatives
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.

### Section 2-2 : Partial Derivatives

8. Find all the 1st order partial derivatives of the following function.

$z = \frac{{{p^2}\left( {r + 1} \right)}}{{{t^3}}} + pr\,{{\bf{e}}^{2p + 3r + 4t}}$ Show Solution

For this problem it looks like we’ll have three 1st order partial derivatives to compute. Here they are,

\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial z}}{{\partial p}} & = {z_p} = \frac{{2p\left( {r + 1} \right)}}{{{t^3}}} + r\,{{\bf{e}}^{2p + 3r + 4t}} + 2pr\,{{\bf{e}}^{2p + 3r + 4t}}\\ \frac{{\partial z}}{{\partial r}} & = {z_r} = \frac{{{p^2}}}{{{t^3}}} + p\,{{\bf{e}}^{2p + 3r + 4t}} + 3pr\,{{\bf{e}}^{2p + 3r + 4t}}\\ \frac{{\partial z}}{{\partial t}} & = {z_t} = - \frac{{3{p^2}\left( {r + 1} \right)}}{{{t^4}}} + 4pr\,{{\bf{e}}^{2p + 3r + 4t}}\end{align*}}

The notation used for the derivative doesn’t matter so we used both here just to make sure we’re familiar with both forms.