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.

Answer and Explanation: 1

Become a Study.com member to unlock this answer!

View this answer

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,\\ ...

See full answer below.


Learn more about this topic:

Loading...
The Chain Rule for Partial Derivatives

from

Chapter 14 / Lesson 4
36K

This lesson defines the chain rule. It goes on to explore the chain rule with partial derivatives and integrals of partial derivatives.


Related to this Question

Explore our homework questions and answers library