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Section 2-2 : Partial Derivatives

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

\[R\left( {x,y} \right) = \frac{{{x^2}}}{{{y^2} + 1}} - \frac{{{y^2}}}{{{x^2} + y}}\] Show Solution

For this problem it looks like we’ll have two 1st order partial derivatives to compute.

Be careful with quotient rules with partial derivatives. For example the first term, while clearly a quotient, will not require the quotient rule for the \(x\) derivative and will only require the quotient rule for the \(y\) derivative if we chose to leave the \({y^2} + 1\) in the denominator (recall we could just bring it up to the numerator as \({\left( {{y^2} + 1} \right)^{ - 1}}\) if we wanted to). The second term on the other hand clearly has \(y\)’s in both the numerator and the denominator and so will require a quotient rule for the \(y\) derivative.

Here are the two 1st order partial derivatives for this problem.

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial R}}{{\partial x}} & = {R_x} = \frac{{2x}}{{{y^2} + 1}} + \frac{{2x{y^2}}}{{{{\left( {{x^2} + y} \right)}^2}}}\\ \frac{{\partial R}}{{\partial y}} & = {R_y} = - \frac{{2y{x^2}}}{{{{\left( {{y^2} + 1} \right)}^2}}} - \frac{{2y{x^2} + {y^2}}}{{{{\left( {{x^2} + y} \right)}^2}}}\end{align*}}\]

Note that we did a little bit of simplification in the derivative work here and didn’t actually show the “first” step of the problem under the assumption that by this point of your mathematical career you can do the quotient rule and don’t really need us to show that step to you.

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