helps with the calc-yoo-luss

HalberdBlue

I just got back from a calculus test and I breezed through it except for one problem that held me up and I still can't figure it out. And since I know I'll dwell on it all day and night until I find out I can't wait until after class tomorrow to ask how its solved :P PA Forums, I choose you!

This is Calculus 1 and so far we've learned stuff up through the chain rule and limits of series. No integration or anything fancy like that.

The problem was:

f(2) = 5
f'(2) = 3
g'(5) = -7

h(x) = g(f(x)). What is h'(2) and why?

I know that (g o f)'(x) = g'(f(x))f'(x) however I don't know how that helps when I'm given g'(5) and not g'(2).

Since the rest of the test took me about 15 minutes I spent the remaining 40 minutes coming up with this horrible convoluted explanation that it equals -21, but my reasoning involved integration and besides that it was completely wrong (but I didn't come up with this reason until 5 minutes before the end of the test so I didn't have time to come up with anything better).

elkna

Heh, just did this today in my class as well. What you want to do is plug in x for the inside of g'(f(x)) first.

Since x = 2,

g'(f(2))*f'(2)

Given the numbers, that would give you this:

g'(5)*(3)

because f(2) = 5 and f'(2) = 3. At this point, it's where the g'(5) value comes in. So:

(-7)(3)

which = -21.

Hope I explained it clearly enough. ;-)

Ark

h'(x) = g'(f(x))*f'(x)

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

h'(2) = g'(5)*3 = -21

I think you might have misread the problem or something, it sounds like you understood how to do it. And hey, you got the right answer anyway. ;-)

HalberdBlue

Yup, as usual, the answer is blindingly obvious once shown to me and makes me feel dumb :P I hate knowing the correct answer yet not knowing how I got it. Maybe I'll get a couple of points despite my absolutely bogus explanation. Thanks!