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Section 2-2 : Partial Derivatives

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

\[w = \cos \left( {{x^2} + 2y} \right) - {{\bf{e}}^{4x - {z^{\,4}}y}} + {y^3}\] Show Solution

This function isn’t written explicitly with the \(\left( {x,y,z} \right)\) part but it is (hopefully) clearly a function of \(x\), \(y\) and \(z\) and so we’ll have three 1st order partial derivatives and each of them should be pretty easy to compute.

Just remember that when computing each individual derivative that the other variables are to be treated as constants. So, for instance, when computing the \(x\) partial derivative all \(y\)’s and \(z\)’s are treated as constants. This in turn means that, for the \(x\) partial derivative, the third term is considered to be a constant (it doesn’t contain any \(x\)’s) and so differentiates to zero. Dealing with these types of terms properly tends to be one of the biggest mistakes students make initially when taking partial derivatives. Too often students just leave them alone since they don’t contain the variable we are differentiating with respect to.

Be careful with chain rule. Again, one of the biggest issues with partial derivatives is students forgetting the “rules” of partial derivatives when it comes to differentiating the inside function. Just remember that you’re just doing the partial derivative of a function and remember which variable we are differentiating with respect to.

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

\[\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}\frac{{\partial w}}{{\partial x}} & = {w_x} = - 2x\sin \left( {{x^2} + 2y} \right) - 4{{\bf{e}}^{4x - {z^{\,4}}y}}\\ \frac{{\partial w}}{{\partial y}} & = {w_y} = - 2\sin \left( {{x^2} + 2y} \right) + {z^4}{{\bf{e}}^{4x - {z^{\,4}}y}} + 3{y^2}\\ \frac{{\partial w}}{{\partial z}} & = {w_z} = 4{z^3}y{{\bf{e}}^{4x - {z^{\,4}}y}}\end{align*}}\]

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