Section 13.3 : Interpretations of Partial Derivatives
1. Determine if \(f\left( {x,y} \right) = x\ln \left( {4y} \right) + \sqrt {x + y} \) is increasing or decreasing at \(\left( { - 3,6} \right)\) if
- we allow \(x\) to vary and hold \(y\) fixed.
- we allow \(y\) to vary and hold \(x\) fixed.
So, we want to determine the increasing/decreasing nature of a function at a point. We know that this means a derivative from our basic Calculus knowledge. Also, from the problem statement we know we want to allow \(x\) to vary while \(y\) is held fixed. This means that we will need the \(x\) partial derivative.
The \(x\) partial derivative and its value at the point is,
\[{f_x}\left( {x,y} \right) = \ln \left( {4y} \right) + \frac{1}{2}{\left( {x + y} \right)^{ - \,\,\frac{1}{2}}}\hspace{0.25in} \to \hspace{0.25in}{f_x}\left( { - 3,6} \right) = \ln \left( {24} \right) + \frac{1}{{2\sqrt 3 }} = 3.4667\]So, we can see that \({f_x}\left( { - 3,6} \right) > 0\) and so at \(\left( { - 3,6} \right)\) if we allow \(x\) to vary and hold \(y\) fixed the function will be increasing.
b allow \(y\) to vary and hold \(x\) fixed Show Solution
This part is pretty much the same as the previous part. The only difference is that here we are allowing \(y\) to vary and we’ll hold \(x\) fixed. This means we’ll need the \(y\) partial derivative.
The \(y\) partial derivative and its value at the point is,
\[{f_y}\left( {x,y} \right) = \frac{x}{y} + \frac{1}{2}{\left( {x + y} \right)^{ - \,\,\frac{1}{2}}}\hspace{0.25in} \to \hspace{0.25in}{f_y}\left( { - 3,6} \right) = - \frac{1}{2} + \frac{1}{{2\sqrt 3 }} = - 0.2113\]So, we can see that \({f_y}\left( { - 3,6} \right) < 0\) and so at \(\left( { - 3,6} \right)\) if we allow \(y\) to vary and hold \(x\) fixed the function will be decreasing.
Note that, because of the three dimensional nature of the graph of this function we can’t expect the increasing/decreasing nature of the function in one direction to be the same as in any other direction!