Two questions:
#1:
(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)=
#1 I'd like to help, but I'm not sure what the notation (fog)'(3) means. Is the ' a prime, representing a derivative?
#2 Find F'(a):
Since F(x) = f(x^6), taking the derivative, we need to use chain rule:
F'(x) = f'(x^6)*6x^5
F'(a) = f'(a^6)*6a^5
Now, we know f'(a^{6})=10, a^{5}=10, so we get
F'(a) = {10}*6*{10} = 600
(fog)(3) = f(g(3))
First we need the value of g at x=3: g(3) = 4. To get this, just look on the x-axis where x=3 and move up until you hit the graph of g, the y-value at that point is g(3)=4.
Second, we need f(g(3)) = f(4) = 2. Use the same method for x=4, going up to the graph of f. The y-value at that point is f(4)=2.
Now, do the same for (gof)(3) = g(f(3)).
The technical rule is that any time you 'take the derivative of y' you get a y'. For example, let's say I'm taking the derivative of 2xy implicitly. This is product rule, so: D(2xy) = 2x*y' + 2y. When I took the derivative of y, I got y', when I took the derivative of x, 1.
Besides the technical rule, what you have done is correct.
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.
Thanks!