Chapter 4: Partial Differentiation

$\mathbf{If}\mathbf{}{\mathbf{xe}}^{\mathbf{y}}\mathbf{=}{\mathbf{ye}}^{\mathbf{x}}\mathbf{,}\mathbf{}\mathbf{find}\mathbf{}\mathbf{dy}\mathbf{/}\mathbf{dx}\mathbf{,}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{d}}^{\mathbf{2}}\mathbf{y}\mathbf{/}{\mathbf{dx}}^{\mathbf{2}}\mathbf{}\mathbf{for}\mathbf{}\mathbf{y}\mathbf{\ne }\mathbf{1}\mathbf{.}$

Given $\mathit{z}\mathbf{=}{\mathit{y}}^{\mathbf{2}}$- $\mathbf{2}{\mathit{x}}^{\mathbf{2}}$, Find ${\left(\frac{\partial z}{\partial x}\right)}_{\mathbf{r}}\mathbf{,}{\left(\frac{\partial z}{\partial \theta }\right)}_{\mathbf{x}}\mathbf{,}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{x}\mathbf{\partial }\mathbf{\theta }}$.

For $\mathit{w}\mathbf{=}\mathbf{8}{\mathit{x}}^{\mathbf{4}}\mathbf{+}{\mathit{y}}^{\mathbf{4}}\mathbf{-}\mathbf{2}\mathit{x}{\mathit{y}}^{\mathbf{2}}$, find${\mathbf{\partial }}^{\mathbf{2}}\mathit{w}\mathbf{/}\mathbf{\partial }{\mathit{x}}^{\mathbf{2}}$and${\mathbf{\partial }}^{\mathbf{2}}\mathit{w}\mathbf{/}\mathbf{\partial }{\mathit{y}}^{\mathbf{2}}$at the points where$\mathbf{\partial }\mathit{w}\mathbf{/}\mathbf{\partial }\mathit{x}\mathbf{=}\mathbf{\partial }\mathit{w}\mathbf{/}\mathbf{\partial }\mathit{y}\mathbf{=}\mathbf{0}$.

To find the maximum and the minimum points of the given function.

$\mathbf{4}\mathbf{+}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}\mathbf{xy}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{y}}^{\mathbf{2}}$

If we are given $\mathit{z}\mathbf{=}\mathit{z}\left(x,y\right)$and $\mathit{y}\mathbf{=}\mathit{y}\left(x\right)$, show that the chain rule (5.1) gives$\frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{y}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$$\frac{\mathbf{d}\mathbf{z}}{\mathbf{d}\mathbf{x}}\mathbf{=}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{y}}\frac{\mathbf{d}\mathbf{y}}{\mathbf{d}\mathbf{x}}$

Let Rbe the resistance of${\mathbf{R}}_{\mathbf{1}}\mathbf{=}\mathbf{25}\mathbf{}\mathbf{ohms}$and${\mathbf{R}}_{\mathbf{2}}\mathbf{=}\mathbf{15}\mathbf{}\mathbf{ohms}$ohms in parallel. (See Chapter 2, Problem 16.6.) If$\mathbf{}{\mathbf{R}}_{\mathbf{1}}$is changed to$\mathbf{}\mathbf{25}\mathbf{.}\mathbf{1}\mathbf{}\mathbf{ohms}$, find$\mathbf{}{\mathbf{R}}_{\mathbf{2}}\mathbf{}$so that$\mathbf{R}$is not changed.

5. If and $\mathit{u}\mathbf{=}{\mathit{x}}^{\mathbf{2}}{\mathit{y}}^{\mathbf{3}}\mathit{z}$,$\mathit{x}\mathbf{=}\mathbf{\text{sin}}\left(s+t\right)\mathbf{,}\mathit{y}\mathbf{=}\mathbf{\text{cos}}\left(s+t\right)$,$\mathit{z}\mathbf{=}{\mathit{e}}^{\mathbf{s}\mathbf{t}}$find$\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{s}}$and$\frac{\mathbf{\partial }\mathbf{u}}{\mathbf{\partial }\mathbf{t}}$.

Find the point on $\mathbf{2}\mathit{x}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{+}\mathit{Z}\mathbf{-}\mathbf{11}\mathbf{=}\mathbf{0}$for which $\mathbf{4}{\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}$is a minimum.

Find the two-variable Maclaurin series for the following functions.

$\sqrt{\mathbf{1}\mathbf{+}\mathbf{xy}}$

If $\mathit{z}\mathbf{=}{\mathit{r}}^{\mathbf{2}}\mathbf{-}{\mathit{x}}^{\mathbf{2}}$, find${\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{r}}\mathbf{\right)}}_{\mathbf{\theta }}\mathbf{,}{\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{\theta }}\mathbf{\right)}}_{\mathbf{r}}\mathbf{,}\frac{{\mathbf{\partial }}^{\mathbf{2}}\mathbf{z}}{\mathbf{\partial }\mathbf{r}\mathbf{\partial }\mathbf{\theta }}\mathbf{,}{\mathbf{\left(}\frac{\mathbf{\partial }\mathbf{z}}{\mathbf{\partial }\mathbf{x}}\mathbf{\right)}}_{\mathbf{y}}$