The total derivative gives the overall dependency of y on x3. Write y = f(x1, x2) = x1*x2, with x1 and x2 depends on x3 through the equation x1+x2+x3=1 Then dy/dx3 = əf/əx1 * əx1/əx3 + əf/əx2 * əx2/əx3 Now əf/əx1 = x2 and əf/əx2 = x1, and əx1/əx3 = -1, əx2/əx3 = -1.Therefore dy/dx3 = –x1 – x2 = x3-1 So a19840’s answer is correct.
To shuangfu:
x1+x2+x3 =1 can be rewritten as: x1 = g(x2, x3) = 1-x2-x3 and x2 = h(x1, x3) = 1-x1-x3
So əx1/əx3 and əx2/əx3 are well defined. With y=f(x1, x2, x3) = x1*x2, where x1 = g(x2, x3) and x2 = h(x1, x3) defined as above, using chain rule, dy/dx3 can be derived as in my original post.
qffq is right in general, but I think that It's better to consider the plan f(x1,x2)=x1+x2=1-x3=h(x3)=f(x1(x3),x2(x3))
So df/dx3=( əf/əx1 * əx1/əx3, əf/əx2 * əx2/əx3)=(-1,-1)
We suppose g(x1,x2,x3)=x1*x2. Since əg/əx3=0,
dg/dx3=dg/df*df/dx3= (x2,x1) * transfom(-1,-1)=-x1-x2=1-x3