# 2 Questions Derivative

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• Mar 21st 2007, 06:25 PM
ecMathGeek
Quote:

Originally Posted by qbkr21
Does anyone know how to solve question #1 with the graph? After studying the picture for quite some time I have come to the conclusion that the lines on the graph are f(x) and g(x), they are not f'(x) and g'(x). Any assistance would wonderful. This is my last problem for Webwork and I am clueless as to what I should do.:confused:

Thanks!

If you can adidquately define (fog)'(3), I can answer the question. I'm not sure what to do with the prime. Does that mean 'take the derivative of (fog)' or does that mean 'the derivative of f OF the derivative of g'? Or does it have nothing to do with derivatives at all? Until I know which is the case, there's not much I can do to help.
• Mar 21st 2007, 06:47 PM
qbkr21
Re:
(fog)'(x) = f'(g(x))*g'(x))
• Mar 21st 2007, 07:03 PM
ecMathGeek
Quote:

Originally Posted by qbkr21
Two questions:

#1:

http://item.slide.com/r/1/118/i/wVF9...lyy5jfoo0d5sx/

(fog)'(3)=?
(gof)'(3)=?

#2:

Let F(x)= f(x^{6}) and G(x)=(f(x))^{6} . You also know that a^{5}=10, f(a)=2,f'(a)=12, f'(a^{6})=10.

Find F'(a)=

(fog)'(3) = f'(g(3))g'(3)
g(3) = 4 ... just go to x = 3, then move up until you hit the graph of g(x), that y value is the value of g(3), which is 4.
f'(g(3)) = f'(4) = 2 ... go to x = 4, then up to the graph of f(x), estimate the slope of f(x) at that point and you will have the value of f'(4), which is 2.
g'(3) = -1 ... explanation: same as above
(fog)'(3) = (2)(-1) = -2

(gof)'(3) = g'(f(3))f'(3)
f(3) = 0
g'(f(3)) = g'(0) = 0
f'(3) = 2
(gof)'(3) = (0)(2) = 0
• Mar 21st 2007, 07:05 PM
qbkr21
Re:
BINGO!:)

1241Sec3.5: Problem 11