# Find h'(2)? (see image below)

Mar 14, 2017

The graphs above were:

Some things to keep in mind:

• Note that $h \left(x\right) = f \left(g \left(x\right)\right)$. This means you will have to find a value for $g \left(x\right)$, and use that value for the argument of $f \left(x\right)$. For instance, for some $g \left(3\right) = w$, we have that $h \left(3\right) = f \left(g \left(3\right)\right) = f \left(w\right)$.
• We should recognize that the derivative at a corner, e.g. when the graph abruptly changes slope, is undefined.
• $h ' \left(x\right) = f ' \left(g \left(x\right)\right) \cdot g ' \left(x\right)$ by the chain rule.

Thus, $h ' \left(x\right)$ does not exist if $g \left(x\right)$ or $g ' \left(x\right)$ do not exist. (If $f ' \left(x\right)$ does not exist at some specified point, it doesn't necessarily imply that $g \left(x\right)$ does not exist.)

For $h ' \left(2\right)$, we have that:

$h ' \left(2\right) = f ' \left(g \left(2\right)\right) g ' \left(2\right)$

$= f ' \left(4\right) g ' \left(2\right)$

$= 3 \cdot - 1$

$= - 3$

Therefore, $h ' \left(2\right) = - 3$, and it exists. Two of the question options apparently has us check $h ' \left(- 2\right)$, so let's see.

$h ' \left(- 2\right) = f ' \left(g \left(- 2\right)\right) g ' \left(- 2\right)$

$= f ' \left(1\right) g ' \left(- 2\right)$

$= - 1 \cdot - 4$

$= 4$

Well, we found that $h ' \left(2\right)$ exists... but neither of the answer choices in which it exists has the correct $h ' \left(- 2\right)$. Either there is a typo in the question or there is no correct answer choice.