Question

Question on notation!

What does (fog)' represent?

Answer 1

(fof) = f(g(x))
solve g(x) and plug the solution to g(x) into f(x) as the x value.

Answer 2

Thanks tomo!

I understand that.
I was wondering how the derivative works for this
(fog)'

Answer 3

ok, so say g(x) = x^3 and f(g(x)) = x^2

plug x^3 from g(x) into f(g(x)) = (x^3)^2. take the derivative of f(g(x)) then.

Answer 4

@tomo
So would this be


?

We're looking at the change with respect to what when we do the overall derivative?

Answer 5

i am pretty sure that it is the second one.

Answer 6

so by (fog)' we'd be looking for d/dg(x) then?
I am still trying to understand what we're looking for the change of through composite (fog)-like functions.
The difference in notation gets me slightly confused.

How is the notation between the ' and the d/dx interchanged?

Answer 7

\[(f\circ g)'=f'(g)g'\] that is the chain rule

Answer 8

for example if \(f(x)=\sin(x)\) and \(g(x)=x^3\) then
\[f\circ g(x)=\sin(x^3)\] and
\[(f\circ g)'=\cos(x^3)\times 3x^2\]

Answer 9

@satellite73
Is \[(f o g)' \] meaning the change in f(g(x)) as g(x) changes?

Let's say we have \[(f o g)'(x)\]
What would that mean in writing?
What essentially is happening here?

Answer 10

It's the rate of change of fog(x) wrt x.

Answer 11

Meaning,
We're seeing how f(g(x)) is changing as x is changing, correct?

Answer 12

Correct. I think you were over-complicating it.

Answer 13

@agent0smith
Perhaps (:

So that would also mean that...
(fog)'(h(x))

Would mean like:

d/dh(x) (f(g(h(x))))

Correct?
I really appreciate your assistance!

Answer 14

That looks about right...

Answer 15

I appreciate it! (: