I'm having trouble trying to visualize graphically the Chain Rule.
When finding the derivative of a function $f$ with respect to $x$ at some point $x = a$ we write $f'(a)$, and this notation denotes that we are looking for the instant rate of change of $f$ at $x = a$. In other words, we are evaluating $f'$ at $x = a$. Graphically, this would be the tangent line to the curve $f$ at $x = a$.
For the Chain Rule, we have
$$f'(g(x)) \cdot g'(x)$$
I think the derivative of $g$ is pretty straight forward since it's being differentiated with respect to $x$. It's the $f'$ that I'm having trouble visualizing graphically since it's being evaluated at the output of $g(x)$ instead of $x$.
When $x = a$, then $g(a)$ which means $f'(g(a))$. So the tangent line to $f$ is occurring at a point $g(a)$? But on which axis is the point $g(a)$ occurring, is it the $x$-axis or the $y$-axis or maybe some other third axis? I'm not sure how to interpret this graphically since I'm only familiar with using $x = a$ as the point at which a tangent line occurs to a curve.
$f'(g(a))$ is the slope of $f(x)$ at $x=g(a)$
However say I have:
$h(x) = f(g(x))$
Then
$h'(x) = f'(g(x))g'(x)$ (this is the chain rule)
So if I want the slope of $h(x)$ at $x=a$ that would be $h'(a) = f'(g(a))g'(a)$
Try this with an example. Say
$$f(x)=x^2$$
$$g(x) = x+3$$
$$h(x) = f(g(x)) = (x+3)^2$$
So the $f'(g(a))$ is like an intermediate value I use to then calculate $h'(a)$.
Try calculating $h'(x)$ directly first. Then try calculating $h'(x)$ using the chain rule. Also try graphing and seeing the results. You will see that the slope of f(x) at $x=g(a)$ is $f'(g(a))$