fc51a1…: “From a silver fish”

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Suppose that $f(x,y)= 2x+y^2-4$ , $x(r,s)=2\cos (\pi rs)+4s^2-4$ and $y(r,s) = e^{2r+3s}$ . Find $\pp [f]{r}$ and $\pp [f]{s}$ when $r=2$ and $s=1$ .

When $r=2$ and $s=1$ ,

$\begin{align*} \pp[f]{r} = \answer{4e^{14}} \\ \pp[f]{s} = \answer{16+6e^{14} } . \end{align*}$

To find $\pp [f]{r}$ , we start by applying chain rule

$\pp [f]{r} = \pp [f]{x}\pp [x]{r} + \pp [f]{y}\pp [y]{r}.$

We now compute each derivative.

$\pp [f]{x} = \answer {2}$
$\pp [f]{y} = \answer {2y}$
$\pp [x]{r} = \answer {- 2 \pi s \sin (\pi rs)}$
$\pp [y]{r} = \answer {2e^{2r+3s}}$

Notice that when $r=2$ and $s=1$ , we have that $x(2,1) = \answer {2}$ and $y(2,1) = \answer {e^7}$ . Substituting into the partial derivatives, we also have that

$\begin{align*} \pp[f]{x} &= \answer{2} \\ \pp[f]{y} &= \answer{2e^7} \\ \pp[x]{r} &= \answer{0} \\ \pp[y]{r} &= \answer{2e^7} \end{align*}$