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### Section 13.6 : Chain Rule

10. Compute $$\displaystyle \frac{{\partial z}}{{\partial x}}$$ and $$\displaystyle \frac{{\partial z}}{{\partial y}}$$ for the following equation.

${{\bf{e}}^{z\,y}} + x{z^2} = 6x{y^4}{z^3}$

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Start Solution

First a quick rewrite of the equation.

${{\bf{e}}^{z\,y}} + x{z^2} - 6x{y^4}{z^3} = 0$ Show Step 2

From the rewrite in the previous step we can see that,

$F\left( {x,y} \right) = {{\bf{e}}^{z\,y}} + x{z^2} - 6x{y^4}{z^3}$

We can now simply use the formulas we derived in the notes to get the derivatives.

$\frac{{\partial z}}{{\partial x}} = - \frac{{{F_x}}}{{{F_z}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{{z^2} - 6{y^4}{z^3}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}} = \frac{{6{y^4}{z^3} - {z^2}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}}}}$ $\frac{{\partial z}}{{\partial y}} = - \frac{{{F_y}}}{{{F_z}}} = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{z{{\bf{e}}^{z\,y}} - 24x{y^3}{z^3}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}} = \frac{{24x{y^3}{z^3} - z{{\bf{e}}^{z\,y}}}}{{y{{\bf{e}}^{z\,y}} + 2xz - 18x{y^4}{z^2}}}}}$

Note that in for the second form of each of the answers we simply moved the “-” in front of the fraction up to the numerator and multiplied it through. We could just have easily done this with the denominator instead if we’d wanted to.