Proof of the Derivative of Exponential Functions using Logarithmic Differentiation

Here is some working proof of deriving the exponential functions that I calculate. I know there's a lot on the internet and there is some other way to prove it. Tbh, I didn't really know how the formula is derived as in the figure above. It's like, how did this happen? Maybe I was too unknowledgeable to figure it out or lack of mathematical (but I just did after a few strive). So here is some working proof, if you guys struggling to figure it out. Sometimes, I review my own entries if I start to forget, lol.

#1 Proof for the derivative of \(\displaystyle\frac{d}{dx}e^x=e^x\),
Let, \(\displaystyle y=e^x\)

\(\displaystyle \ln{y}=\ln{e^x}\)
\(\displaystyle \ln{y}=x\ln{e}\,\,\Rightarrow\,\,\ln{e}=1\)
\(\displaystyle \ln{y}=x\cdot 1\)
\(\displaystyle\frac{d}{dx}[\ln{y}]=\frac{d}{dx}[x]\)
\(\displaystyle\frac{1}{y}y'=1\)
\(\displaystyle y'=y\)
Substitute \(\displaystyle y=e^x\),
\(\displaystyle y'=e^x\)
\(\displaystyle \frac{d}{dx}[e^x]=e^x\)                                \(\square\)

#2 Proof for the derivative of \(\displaystyle\frac{d}{dx}a^x=a^x\ln{a}\),
Let, \(\displaystyle y=a^x\)

\(\displaystyle \ln{y}=\ln{a^x}\)
\(\displaystyle \ln{y}=x\ln{a}\,\,\Rightarrow\,\,\ln{e}=1\)
\(\displaystyle \ln{y}=\ln{a}\cdot \frac{d}{dx}[x]\)

\(\displaystyle\frac{1}{y}y'=\ln{a}\cdot 1\)
\(\displaystyle\frac{1}{y}y'=\ln{a}\)
\(\displaystyle y'=y\cdot\ln{a}\)

Substitute \(\displaystyle y=a^x\),
\(\displaystyle y'=a^x\)
\(\displaystyle \frac{d}{dx}[a^x]=a^x\cdot\ln{a}\)                                \(\square\)

#3 Proof for the derivative of \(\displaystyle\frac{d}{dx}e^u=e^u\cdot \frac{d}{dx}[e^u]\),

Let, \(\displaystyle y=e^u\)

\(\displaystyle \ln{y}=\ln{e^u}\)
\(\displaystyle \ln{y}=u\ln{e}\,\,\Rightarrow\,\,\ln{e}=1\)
\(\displaystyle \ln{y}=u\cdot 1\)
\(\displaystyle\frac{d}{dx}[\ln{y}]=\frac{d}{dx}[u]\)
\(\displaystyle\frac{1}{y}y'=\frac{d}{dx}[u]\)
\(\displaystyle y'=y\cdot \frac{d}{dx}[u]\)

Substitute \(\displaystyle y=e^u\),
\(\displaystyle y'=e^u\cdot \frac{d}{dx}[u]\)

\(\displaystyle \frac{d}{dx}[e^u]=e^u\cdot \frac{d}{dx}[u]\)                                \(\square\)

#4 Proof for the derivative of \(\displaystyle\frac{d}{dx}a^u=a^u\ln{a}\frac{d}{dx}[u]\),

(Keeping in mind the Chain Rule and any variable restrictions).

Let, \(\displaystyle y=a^u\)

\(\displaystyle \ln{y}=\ln{a^u}\)
\(\displaystyle \ln{y}=u\ln{a}\)

\(\displaystyle\frac{d}{dx}[\ln{y}]=\frac{d}{dx}[u\ln{a}]\)

\(\displaystyle\frac{1}{y}y'=\ln{a}\cdot\frac{d}{dx}[u]\)
\(\displaystyle y'=\ln{a}\cdot\frac{d}{dx}[u]\)
\(\displaystyle y'=y\ln{a}\cdot\frac{d}{dx}[u]\)

Substitute \(\displaystyle y=a^u\),
\(\displaystyle y'=a^u\cdot\ln{a}\cdot\frac{d}{dx}[u]\)

\(\displaystyle \frac{d}{dx}[a^u]=a^u\cdot\ln{a}\cdot\frac{d}{dx}[u]\)                                \(\square\)

E.g. (For example proof #4).

\(\displaystyle \tan{e^{(y-x)}}\)

Let, \(\displaystyle y=\tan{e^{(y-x)}}\)

\(\displaystyle \ln{y}= \ln{(\tan{e^{(y-x)}})}\)

\(\displaystyle\frac{d}{dx}[\ln{y}]=\frac{d}{dx}\bigg[\ln{(\tan{e^{(y-x)}})}\bigg]\)

\(\displaystyle\frac{1}{y}y'=\ln{(\tan{e})}\frac{d}{dx}\bigg[(y-x)\bigg]\)

\(\displaystyle\frac{1}{y}y'=\ln{(\tan{e})}\cdot\bigg[\frac{d}{dx}(y)-\frac{d}{dx}(x)\bigg]\)

\(\displaystyle y'=y\ln{(\tan{e})}\cdot\bigg[\frac{d}{dx}(y)-1\bigg]\)

Substitute \(\displaystyle y=\tan{e^{(y-x)}}\),

\(\displaystyle y'=\tan{e^{(y-x)}}\cdot\ln{(\tan{e})}\cdot\bigg[\frac{d}{dx}(y)-1\bigg]\)