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))**(**D_{1}f(t,2-3t)(1)+D_{2}f(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))**(**D_{1}f(1,-1)(1)+D_{2}f(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!

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Maintained by
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