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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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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,

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