# Suppose f is a differentiable function of x and y, and g(r, s) = f(5r - s, s^2 - 3r)....

## Question:

Suppose{eq}\displaystyle \ f {/eq} is a differentiable function of{eq}\displaystyle \ x {/eq} and{eq}\displaystyle \ y, {/eq} and

{eq}\ g(r, s) = f(5r - s, \ s^2 - 3r). {/eq} Use the table of values below to calculate {eq}\displaystyle \ g_r(2, 7) {/eq} and{eq}\displaystyle \ g_s(2, 7). {/eq}

{eq}\begin{array}{|c|c|c|c|c|} \hline Point & f& g & f_x & f_y \\ \hline (3, 43) & 5 & 3 & 1 & 6 \\ \hline (2, 7) & 3 & 5& 7& 8 \\ \hline \end{array} {/eq}

## Partial Derivatives; Chain Rule:

The Chain Rule tells us:

{eq}\begin{align*} g_r(r,s)&=f_x(x,y)\cdot x_r(r,s)+f_y(x,y)\cdot y_r(r,s),\\ &\text{and}\\ g_s(r,s)&=f_x(x,y)\cdot x_s(r,s)+f_y(x,y)\cdot y_s(r,s), \end{align*} {/eq}

where

{eq}x=5r - s, \text{ and }y= s^2 - 3r . {/eq}

So for {eq}(r,s)=(2,7) {/eq} we have {eq}(x,y)=(3,43). {/eq}

From the table we need {eq}f_x(3,43) \text{ and }f_y(3,43) {/eq} and by direct calculation we can get the partial derivatives {eq}x_r(2,7), {/eq}

{eq}x_s(2,7), {/eq}

{eq}y_r(2,7), {/eq} and

{eq}y_s(2,7). {/eq}

We evaluate {eq}\ g_r(2, 7) {/eq} and{eq}\ g_s(2, 7), {/eq} by substituting the values of the partial derivatives found above.

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We have

{eq}g(r, s) = f(x,y), {/eq}

where

{eq}x=5r - s, \text{ and }y= s^2 - 3r . {/eq}

So we have:

{eq}\begin{align*} x_r&=5,\\ x_s&=-1,\\ ...