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Section 2-2 : Partial Derivatives

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

\[f\left( {x,z} \right) = {{\bf{e}}^{ - x}}\sqrt {{z^4} + {x^2}} - \frac{{2x + 3z}}{{4z - 7x}}\] Show Solution

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

Be careful with product rules and quotient rules with partial derivatives. For example, the first term, while clearly a product, will only need the product rule for the \(x\) derivative since both “factors” in the product have \(x\)’s in them. On the other hand, the first “factor” in the first term does not contain a \(z\) and so we won’t need to do the product rule for the \(z\) derivative. In this case the second term will require a quotient rule for both derivatives.

Basically, be careful to not “overthink” product/quotient rules with partial derivatives. Do them when required but make sure to not do them just because you see a product/quotient. When you see a product/quotient look at the “factors” of the product/quotient. Do both “factors” have the variable you are differentiating with respect to or not and use the product/quotient rule only if they both do.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial f}}{{\partial x}} & = {f_x} = - {{\bf{e}}^{ - x}}{\left( {{z^4} + {x^2}} \right)^{\frac{1}{2}}} + x\,{{\bf{e}}^{ - x}}{\left( {{z^4} + {x^2}} \right)^{ - \,\,\frac{1}{2}}} - \frac{{29z}}{{{{\left( {4z - 7x} \right)}^2}}}\\ \frac{{\partial f}}{{\partial z}} & = {f_z} = 2{z^3}{{\bf{e}}^{ - x}}{\left( {{z^4} + {x^2}} \right)^{ - \,\,\frac{1}{2}}} + \frac{{29x}}{{{{\left( {4z - 7x} \right)}^2}}}\end{align*}}\]

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 product and 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.