We first define the function C: C(t)=h(f(t,2-3t)).
Therefore C(1)=h(f(1,-1))=h(-2)=5 (just looking up the values).
And now the derivative:
C´(t)=h´(f(t,2-3t))(D1f(t,2-3t)(1)+D2f(t,2-3t)(-3)).
This is a triple composition. The 1 comes from the derivative of t
with respect to t, and the -3 comes from the derivative of 2-3t with
respect to t. And we can get numbers by plugging in t=1:
C´(1)=h´(f(1,-1))(D1f(1,-1)(1)+D2f(1,-1)(-3))=h´(-2)((-5)(1)+(4)(-3))=2(-17)=-34. That's one of the numbers on the
left of the handout, so this answer is probably correct!
Maintained by greenfie@math.rutgers.edu and last modified 2/16/2006.