Problem
Let $ f(\theta) $ be a differentiable function.
Let $ g(x) = f(\theta + ax) $.
Can we show that $ g'(0) = \left[ \frac{dg}{dx} \right]_{x=0} = a \frac{df}{d\theta} $?
Example
It appears to be true when I pick a specific example of $ f(\theta) $. For example take $ f(\theta) = \sqrt{1 + \theta^2} $. Then $ g(x) = \sqrt{1 + (\theta + ax)^2} $.
Then $ g'(x) = \frac{(\theta + ax)a}{\sqrt{1 + (\theta + ax)^2}} $. So $ g'(0) = \frac{\theta a}{\sqrt{1 + \theta^2}} $.
Also $ a \frac{df}{d\theta} = a \frac{\theta}{\sqrt{1 + \theta^2}} $.
So we do get $ g'(0) = a\frac{df}{d\theta} $. But is this true in general? If yes, how do we prove this equation?
Attempted Solution
This is what I tried so far.
Let $ u(x) = \theta + ax $. Then $ g(x) = f(u(x)) $. Then $ g'(x) = \frac{dg}{dx} = \frac{df}{du} \frac{du}{dx} = \frac{df}{du} a$.
Now is there a way to show that $ g'(0) = \left[ \frac{df}{du} a \right]_{x=0} = \frac{df}{d\theta} a$?
You see, $f$ is a differentiable function and $\theta$ is the variable representing the input. $g(x) = f(\theta + ax)$ implies that we pick a constant value of $\theta$. Assume $\theta = \theta_0$, We have
$$\ \frac{dg(x)}{dx} = \frac{df(\theta_0 + ax)}{dx} = af'(\theta_0+ax) $$
Plugging in $x=0$,
$$\ g'(0) = af'(\theta_0) = \left [a\frac{df(y)}{dy} \right]_{y=\theta_0} $$
I hope it clarifies now. Note that in your case it was: $$\ \frac{df(u)}{du} \text{ it's the same thing, with just different input variable name.} $$