## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises - Page 984: 16

#### Answer

$g_r(1,2)=-24,g_s(1,2)=28$

#### Work Step by Step

$$g_r(r,s)=f_x(2r-s,s^2-4r)\frac{\partial}{\partial r}(2r-s)+f_y(2r-s,s^2-4r)\frac{\partial }{\partial r}(s^2-4r)=f_x(2r-s,s^2-4r)\cdot2+f_y(2r-s,s^2-4r)\cdot(-4)=2f_x(2r-s,s^2-4r)-4f_y(2r-s,s^2-4r)$$ $$g_s(r,s)=f_x(2r-s,s^2-4r)\frac{\partial}{\partial s}(2r-s)+f_y(2r-s,s^2-4r)\frac{\partial }{\partial s}(s^2-4r)=f_x(2r-s,s^2-4r)\cdot(-1)+f_y(2r-s,s^2-4r)\cdot2s=-f_x(2r-s,s^2-4r)+2sf_y(2r-s,s^2-4r)$$ Notice that for $r=1$ and $s=2$ we have $f_x(2r-s,s^2-4r)=f_x(0,0)$ and $f_y(2r-s,s^2-4r)=f_y(0,0)$. $$g_r(1,2)=2f_x(0,0)-4f_y(0,0)=2\cdot4-4\cdot8=-24$$ $$g_s(1,2)=-f_x(0,0)+2\cdot2\cdot f_y(0,0)=-4+4\cdot8=28$$

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