Question
Let H=F(x,y) and x=g(s,t), y=k(s,t) be differentiable functions. Now suppose that g(1,0)=8, k(1,0)=4, gs(1,0)=8, gt(1,0)=2, ks(1,0)=1,...
Let H=F(x,y) and x=g(s,t), y=k(s,t) be differentiable functions. Now suppose that g(1,0)=8, k(1,0)=4, gs(1,0)=8, gt(1,0)=2, ks(1,0)=1, kt(1,0)=5, F(1,0)=9, F(8,4)=3, Fx(1,0)=13, Fy(1,0)=7, Fx(8,4)=9, Fy(8,4)=2. Find Hs(1,0), that is, the partial derivative of H with respect to s, evaluated at s=1 and t=0.
Answers
Ha flouy) Ho = fx xo + fy Yo [by Chain Rule ] H3 (1,0) = Bc 18,40 gs (1,0) + fy (8140 ks (10) = 9 (8) + 2 (1) = 12+2 = 74 I fx 1814)=9 fj 1814 1=2 9s (1,01=8 a (1,0)=1
![Ha flouy) Ho = fx xo + fy Yo [by Chain Rule ] H3 (1,0) = Bc 18,40 gs (1,0) + fy (8140 ks (10) = 9 (8) + 2 (1) = 12+2 = 74 I f](https://i.imgur.com/d4uIoGX.png)
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