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Expert-verified Found in: Page 656 ### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280 # Determine the values of the derivatives $${g_r}(1,2)$$and $${g_r}(1,2).$$The functions are $$g(r,s) = f\left( {2r - s,{s^2} - 4r} \right),x = x(r,s)$$ and $$y = y(r,s).$$

The values of derivatives are $${g_r}(1,2) = - 24$$and$${g_s}(1,2) = 28$$.

See the step by step solution

## Step 1: The chain rule.

With the chain rule, the partial derivative of function $${\rm{f(x, y)}}$$ with respect to $${\rm{t}}$$ becomes $$\frac{{\partial f}}{{\partial t}} = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial t}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial t}}.$$

## Step 2: Use the chain rule for calculation.

Differentiate $$g$$ partially with respect to $$r$$ as follows.

\begin{aligned}{l}\frac{{\partial g}}{{\partial r}} = \frac{\partial }{{\partial r}}f(x(r,s),y(r,s))\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial r}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial r}}\\ = {f_x}(x(r,s),y(r,s)) \cdot {x_r}(r,s) + {f_y}(x(r,s),y(r,s)) \cdot {y_r}(r,s)\\ = {f_x}(x(r,s),y(r,s)) \cdot 2 + {f_y}(x(r,s),y(r,s)) \cdot ( - 4)\end{aligned}

At $$r = 1$$and$$s = 2$$, the partial derivatives becomes,

\begin{aligned}{l}{\left. {\frac{{\partial g}}{{\partial r}}} \right|_{r = 1,s = 2}} = 2{f_x}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right) - 4{f_y}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right)\\ = 2{f_x}(0,0) - 4{f_y}(0,0)\\ = 2(4) - 4(8)\\ = - 24\end{aligned}

That is, $${g_r}(1,2) = - 24$$.

Differentiate $$g$$ partially with respect to $$s$$ as follows.

\begin{aligned}{l}\frac{{\partial g}}{{\partial s}} = \frac{\partial }{{\partial s}}f(x(r,s),y(r,s))\\ = \frac{{\partial f}}{{\partial x}} \cdot \frac{{\partial x}}{{\partial s}} + \frac{{\partial f}}{{\partial y}} \cdot \frac{{\partial y}}{{\partial s}}\\ = {f_x}(x(r,s),y(r,s)) \cdot {x_s}(r,s) + {f_y}(x(r,s),y(r,s)) \cdot {y_s}(r,s)\\ = {f_x}(x(r,s),y(r,s)) \cdot ( - 1) + {f_y}(x(r,s),y(r,s)) \cdot (2s)\end{aligned}

At $$r = 1$$ and$$s = 2$$, the partial derivatives becomes,

\begin{aligned}{l}{\left. {\frac{{\partial g}}{{\partial s}}} \right|_{r = 1,s = 2}} = - {f_x}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right) + 2(2){f_y}\left( {2(1) - (2),{{(2)}^2} - 4(1)} \right)\\ = - {f_x}(0,0) + 4{f_y}(0,0)\\ = - (4) + 4(8)\\ = 28\end{aligned}

That is, $${g_s}(1,2) = 28$$.

Therefore, the values of derivatives are $${g_r}(1,2) = - 24$$and $${g_s}(1,2) = 28$$. ### Want to see more solutions like these? 