It's been awhile since I've seen this kind of notation, so I'm going to assume that ' denotes the inverse function. As in g'(g(x)) = g(g'(x)) = x.
I believe then that the answer is no. Again, if I remember correctly, (f∘g)(x) = f(g(x)). You can assign a third function, h(x), such that h(x) = (f∘g)(x). Then h'(x) = (f∘g)'(x). Since h(h'(x)) = x, and h(x) = f(g(x)), then f(g((f∘g)'(x))) = x. To "cancel out" the functions, (f∘g)'(x) has to be g'(f'(x)). Thus f(g((f∘g)'(x))) becomes f(g(g'(f'(x)))), which is f(f'(x)) which is x.
So the solution would be (f∘g)'(x) = g'(f'(x)).
If you just wanted to find out whether or not (f∘g)'(x) = f(g'(x)), then you could just do the obvious test:
You know that
(f∘g)(f∘g)'(x) = x
So is this true?
f(g(f(g'(x)))) = x
Since it does not appear to be always true for any valid functions f and g, you must claim it to be false.
Comments
It's been awhile since I've seen this kind of notation, so I'm going to assume that ' denotes the inverse function. As in g'(g(x)) = g(g'(x)) = x.
I believe then that the answer is no. Again, if I remember correctly, (f∘g)(x) = f(g(x)). You can assign a third function, h(x), such that h(x) = (f∘g)(x). Then h'(x) = (f∘g)'(x). Since h(h'(x)) = x, and h(x) = f(g(x)), then f(g((f∘g)'(x))) = x. To "cancel out" the functions, (f∘g)'(x) has to be g'(f'(x)). Thus f(g((f∘g)'(x))) becomes f(g(g'(f'(x)))), which is f(f'(x)) which is x.
So the solution would be (f∘g)'(x) = g'(f'(x)).
If you just wanted to find out whether or not (f∘g)'(x) = f(g'(x)), then you could just do the obvious test:
You know that
(f∘g)(f∘g)'(x) = x
So is this true?
f(g(f(g'(x)))) = x
Since it does not appear to be always true for any valid functions f and g, you must claim it to be false.
yes:
imagine X=2;g=3;f=4 no matter how u plug it in it all ends up multiplied.