Why the partial derivative for these two similar cases are done differently? I was watching patrickJMT's derivates of logarithmic functions video and I got stuck understanding why these derivates are treated differently:
$f(x) = \ln(g(x))$ derivative is $f'(x) = \frac{1 }{ g(x)} * g'(x)$
while the function $f(x) = log_a g(x)$  has the derivative $f'(x) = \frac{1 }{ g(x) \ln(a)}$.
Why do we not include the derivative of $g(x)$ as well to the case of log function?
Reference
 A: You are correct: the derivative of $\log_a(g(x))$ is in fact $\displaystyle{\frac{g'(x)}{g(x)\ln a}}$. This is because, by the chain rule
$$
\frac{d}{dx}\left(\log_a(g(x))\right)=(\log_a)'(g(x))\cdot g'(x)=\frac{1}{u\ln a}\bigg\rvert_{u=g(x)}\cdot g'(x)=\frac{g'(x)}{g(x)\ln a} \, .
$$
Note that since $f$ and $g$ are functions of one variable, their partial derivatives are exactly the same as their total derivatives. The concept of a "partial derivative" is only useful when considering functions that depend on several variables, such as $f(x,y)=xy+y^3$.
A: You are right that the chain rule should give a $g'(x)$ factor in both derivatives. The only difference should be the $\ln(a)$ factor in the logarithm to base $a$ case.
It's also worth pointing out that they are both derivatives of logarithms. $ln(x)=\log_e(x)$, which explains why there is no extra factor since $\ln(e)=1$.
A: Recall that
$$\log_A B= \frac{\log B}{\log A} \implies \log_a g(x)= \frac{\log g(x)}{\log a}$$
and therefore
$$(\log_a g(x))'= \frac{(\log g(x))'}{\log a}=\frac{g'(x)}{g(x) \log a}$$
