884673…: “Beside a sad calculation”

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Suppose that $f(x,y)$ is differentiable, $\grad {f}(5,2) = \vector {-1,2}$ , $x(r,s)=2r+3s$ and $y(r,s) = 2r^2s$ .

Is there enough information to compute $\pp [f]{s}$ when $r=1$ and $s=1$ ?

Yes. No.

In order to compute $\pp [f]{s}$ when $r=1$ and $s=1$ , we need $f_x(x,y)$ and $f_y(x,y)$ at the $x$ and $y$ - values corresponding to $r=1$ and $s=1$ . Noting that $x(1,1) = 5$ and $y(1,1) = 2$ , this means we need $f_x(5,2)$ and $f_y(5,2)$ , which we have.

Find $\pp [f]{s}$ when $r=1$ and $s=1$ .

When $r=1$ and $s=1$ , we find

$\pp [f]{s} = \answer {1}.$

We can find $f_x(5,2)$ and $f_y(5,2)$ from $\grad {f}(5,2)$ .