Trying to understand the following proof of $f(x) = a^x \implies f'(x) = a^x \ln a$

Question

I was reading proof of how to obtain derivative of $a^x$ here (Click on the screenshot below to expand).

Although I comprehend the proof, I have several questions about the way the proof is constructed and communicated:

1. Why did he substitute $u$ for $x\cdot \ln a$? He could've just directly used the chain rule without doing any substitution. (My only guess would be that this step was for clarification purpose, yet I fail to see how this makes things clearer)
2. What does phrase "Cancel out $dx$ over $dx$" mean? I understand what he did right after saying this, but the phrase itself perplexes me.

Answer 1

There is no need at all to introduce a new variable $u = x \ln(a)$ if you know how to make use of the chain rule. All you're doing is writing things as a power of $e$. So, $f(x) = a^x = e^{x \ln(a)}$. Hence, by the rule for differentiating exponentials, and by the chain rule, \begin{align} f'(x) = \ln(a) \cdot e^{x \ln(a)} = \ln(a) \cdot a^x \end{align}

Once you know how to apply the chain rule, there is no need at all to explicitly introduce new variables (or more strictly speaking, write out all the compositions involved).

For the second question, that is just a very weird, and I would even say misleading way of saying that $\dfrac{dx}{dx} = 1$, because you're not "cancelling" the $dx$'s. The most formal way to state it is that the derivative of the identity function $I:\Bbb{R} \to \Bbb{R}$, defined by $I(x) = x$, is the constant function $1$; i.e for every $x \in \Bbb{R}$, $I'(x) = 1$.

It seems their "added" explanation just confused you more so, my suggestion is to simply ignore it, and phrase the argument in your own words so that you understand it better.

Answer 2

Before answering your questions, I'll just note the $\frac{\mathrm{d}}{\mathrm{d}x}e^u$ factor should read $\frac{\mathrm{d}}{\mathrm{d}u}e^u$.

1. It is just for clarification. We have $a^x=f(g(x))$ with $f(u)=e^u,\,g(x)=x\ln a$.
2. This means $$\frac{\mathrm{d}}{\mathrm{d}x}\left(x\ln a\right)=\left(\frac{\mathrm{d}}{\mathrm{d}x}x\right)\ln a=\frac{\mathrm{d}x}{\mathrm{d}x}\ln a=1\cdot\ln a=\ln a.$$

Answer 3

For 1, the issue is that the next formula is wrong. You have $\frac d{dx}e^u$. It should be instead $\frac d{du}e^u$. For 2, it takes $\frac d{dx}x\ln a$ and they write it as $\ln a\frac{dx}{dx}$. It's a bad way of saying that the derivative of $x$ with respect to $x$ is $1$.