Why is derivate of log(u) with respect to u is not 1/u? I am trying to learn derivates again. And was going over this calculation
https://www.mathway.com/popular-problems/Calculus/531323
It confused me :

I thought derivative of log(x) with respect to x is 1/x.
Source: https://simple.wikipedia.org/wiki/Derivative_(mathematics)
 A: Mathematicians writing $\log u$ almost always mean the base-$e$ logarithm of $u$, and then one has $$\frac d {du} \log u = \frac 1 u.$$ Calculators almost always take $\log$ to mean base-$10$ logarithm, and one has $$\frac d {du} \log_{10} u = \frac 1 {u \log_e 10}.$$ The convention that $\log$ means the base-$10$ logarithm is followed in some contexts in some disciplines in science and engineering, but is not generally followed in mathematics.
A: If your $\log x$ is $\ln x,$ then what you said is correct.
However, some regard $\log x$ as $\log_{10} x.$ In this case you will have
$$(\log u)'=\left(\frac{\ln u}{\ln 10}\right)'=\frac{1}{u\ln 10}.$$
I think since in your reference, they had used $\ln$ to denote natural log, then $\log$ may be used to mean above.
A: You're confusing the derivative of the natural logarithm, and the logarithm in base $10$. For the natural logarithm we have:
$$\frac{d}{du}\log_e(u)=\frac{d}{du}\ln(u)=\frac{1}{u}\qquad u> 0$$
Now for the base $10$ logarithm (often simply denoted $\log(u)$), we use the base conversion formula to express the logarithm in terms of the natural logarithm (base $e$), and then we differentiate. We have:
$$\frac{d}{du}\log_{10}(u)=\frac{d}{du}\log(u)=\frac{d}{du}\frac{\ln(u)}{\ln(10)}=\frac{1}{\ln(10)u}\qquad u> 0$$
