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

##### 1 Answer

Mar 14, 2017

The graphs above were:

Some things to keep in mind:

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

Thus,

Reading from the above graph:

For

#h'(2) = f'(g(2))g'(2)#

#= f'(4)g'(2)#

#= 3*-1#

#= -3#

Therefore,

#h'(-2) = f'(g(-2))g'(-2)#

#= f'(1)g'(-2)#

#= -1*-4#

#= 4#

Well, we found that