Dr. Z’s Math251 Handout #14.6 (2nd ed.) [The Chain Rule]By Doron ZeilbergerProblem Type 14.6a: Use the chain rule to find∂z∂sand∂z∂t:z = f(x, y) , x = h1(s, t) , y = h2(s, t) .Example Problem 14.6a: Use the chain rule to find∂z∂sand∂z∂t:z = x3+ 2xy + y2, x = s + 2t , y = s2t .Steps Example1. Set-up the chain-rule:∂z∂s=∂z∂x∂x∂s+∂z∂y∂y∂s,∂z∂t=∂z∂x∂x∂t+∂z∂y∂y∂t.1. Ditto.2. For the specific functions given com-pute all the necessary quantities that showup on the right sides, namely:∂z∂x,∂z∂y,∂x∂s,∂y∂s,∂x∂t,∂y∂t.2.∂z∂x= 3x2+ 2y ,∂z∂y= 2x + 2y .∂x∂s= 1 ,∂y∂s= 2st .∂x∂t= 2 ,∂y∂t= s2.3. Incorporate them into the fomulas ofstep 1.3.∂z∂s= (3x2+2y)·1+(2x+2y)(2st) = 3x2+2y+4xst+4yst .∂z∂t= (3x2+2y)·2+(2x+2y)s2= 6x2+4y+2xs2+2ys2.Ans.:∂z∂s= 3x2+ 2y + 4xst + 4yst,∂z∂t=6x2+ 4y + 2xs2+ 2ys2.1Note: It is acceptable to have x, y featuring in the answers as well as t, s. If you are asked toexpress your answers only in terms of the “basic” variables t, s, then you would have to do an extrastep: substituting everywhere you see x or y their expressions in terms of t and s. So in that case,the answers would be∂z∂s= 3(s + 2t)2+ 2s2t + 4(s + 2t)st + 4(s2t)st ,∂z∂t= 6(s + 2t)2+ 4s2t + 2(s + 2t)s2+ 2(s2t)s2.Do not bother to expand (unless specifically asked to).A Problem from a previous FinalFind∂f∂rand∂f∂sas functions of r and s , iff(x, y) = x3+ 2xy + y3,and the variables are related by x = r − s and y = r + s. You do not need to simplify!Ans.∂f∂r= 3(r − s)2+2(r +s)+3(r +s)2+2(r − s),∂f∂s= −3(r − s)2− 2(r +s)+3(r +s)2+2(r − s).Another Problem from a Previous FinalFind∂z∂xand∂z∂yifsin(x + 2y + 3z) = 5xyz + 1 .Ans. (5yz − cos(x + 2y + 3z))/(3 cos(x + 2y + 3z) − 5xy); (5xz − 2 cos(x + 2y + 3z))/(3 cos(x +2y + 3z) − 5xy)
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