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Section 2-2 : Partial Derivatives

9. Find \(\frac{{\partial z}}{{\partial x}}\) and \(\frac{{\partial z}}{{\partial y}}\) for the following function.

\[{x^2}\sin \left( {{y^3}} \right) + x{{\bf{e}}^{3z}} - \cos \left( {{z^2}} \right) = 3y - 6z + 8\]

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Start Solution

Okay, we are basically being asked to do implicit differentiation here and recall that we are assuming that \(z\) is in fact \(z\left( {x,y} \right)\) when we do our derivative work.

Let’s get \(\frac{{\partial z}}{{\partial x}}\) first and that requires us to differentiate with respect to \(x\). Just recall that any product involving \(x\) and \(z\) will require the product rule because we’re assuming that \(z\) is a function of \(x\). Also recall to properly chain rule any derivative of \(z\) to pick up the \(\frac{{\partial z}}{{\partial x}}\) when differentiating the “inside” function.

Differentiating the equation with respect to \(x\) gives,

\[2x\sin \left( {{y^3}} \right) + {{\bf{e}}^{3z}} + 3\frac{{\partial z}}{{\partial x}}x{{\bf{e}}^{3z}} + 2z\frac{{\partial z}}{{\partial x}}\sin \left( {{z^2}} \right) = - 6\frac{{\partial z}}{{\partial x}}\]

Solving for \(\frac{{\partial z}}{{\partial x}}\) gives,

\[2x\sin \left( {{y^3}} \right) + {{\bf{e}}^{3z}} = \left( { - 6 - 3x{{\bf{e}}^{3z}} - 2z\sin \left( {{z^2}} \right)} \right)\frac{{\partial z}}{{\partial x}}\hspace{0.5in} \to \hspace{0.5in} \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\partial z}}{{\partial x}} = \frac{{2x\sin \left( {{y^3}} \right) + {{\bf{e}}^{3z}}}}{{ - 6 - 3x{{\bf{e}}^{3z}} - 2z\sin \left( {{z^2}} \right)}}}}\] Show Step 2

Now we get to do it all over again except this time we’ll differentiate with respect to \(y\) in order to find \(\frac{{\partial z}}{{\partial y}}\). So, differentiating gives,

\[3{y^2}{x^2}\cos \left( {{y^3}} \right) + 3\frac{{\partial z}}{{\partial y}}x{{\bf{e}}^{3z}} + 2z\frac{{\partial z}}{{\partial y}}sin\left( {{z^2}} \right) = 3 - 6\frac{{\partial z}}{{\partial y}}\]

Solving for \(\frac{{\partial z}}{{\partial y}}\) gives,

\[3{y^2}{x^2}\cos \left( {{y^3}} \right) - 3 = \left( { - 6 - 3x{{\bf{e}}^{3z}} - 2zsin\left( {{z^2}} \right)} \right)\frac{{\partial z}}{{\partial y}}\hspace{0.5in} \to \hspace{0.5in} \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{\partial z}}{{\partial y}} = \frac{{3{y^2}{x^2}\cos \left( {{y^3}} \right) - 3}}{{ - 6 - 3x{{\bf{e}}^{3z}} - 2zsin\left( {{z^2}} \right)}}}}\]