How to remember differentiation formula for logarithmic functions So, I was trying to memorize the differentiation formulas for logarithmic functions and exponential functions but It's way too much. I keep forgetting whether to use ln and then multiply the expression by it's power and stuff like that. Is there a clean way to remember these formulas without much pain? Thank you!
 A: Please memorise only these few items; it is straightforward to derive all the remaining formulae using the chain rule:

\begin{align}&\displaystyle\frac{\mathrm d}{\mathrm dx} e^x=e^x &&a^x:=e^{\ln(a^x)}\\\\
  &\displaystyle\frac{\mathrm d}{\mathrm dx} \ln |x|=\frac1x &&\log_ax=\frac{\ln x}{\ln a} \\&\text{(if $x\not<0,$ omit the | | symbol)}\end{align}

For example, $$\frac{\mathrm d}{\mathrm dx} \log_af(x)=\frac{\mathrm d}{\mathrm dx} \left(\frac{\ln f(x)}{\ln a}\right)=\frac1{\ln a}\left(f'(x)\frac1{f(x)} \right)=\frac{f'(x)}{(\ln a)f(x)};\\
\frac{\mathrm d}{\mathrm dx} a^{f(x)}=\frac{\mathrm d}{\mathrm dx} \left(e^{f(x)\ln a}\right)=(\ln a)f'(x)\left(e^{f(x)\ln a}\right)=(\ln a)f'(x)a^{f(x)}.$$
Bonus: $$\frac{\mathrm d}{\mathrm dx} x^{x}=\frac{\mathrm d}{\mathrm dx} \left(e^{x\ln x}\right)=\left(x\left(\frac1x\right)+\ln x\right)e^{x\ln x}=(1+\ln x)x^x.$$
A: Just remember that $(e^x)' = e^x$, all the rest follows from that and the chain rule.
For example, suppose you want to find $(\ln x)'$. Then you just write
$$y = \ln x$$
$$e^y = x\tag{defn of ln}$$
$$(e^y)' = (x)'\tag{diff each side}$$
$$e^y\cdot y' = 1\tag{deriv of $e^x$ + chain rule on LHS}$$
$$y' = \frac1{e^y}\tag{div by $e^y$}$$
$$y' = \frac1x\tag{since $e^y=x$ from 2nd line}$$
The only "hard" thing in any of this is to remember the chain rule, which says $[f(g(x)]' = f'(g(x))\cdot g'(x)$.
The chain rule is used because you aren't differentiating just plain $e^x$, you are differentiating $e^y$, and $y$ is a function of $x$ (in other words, you're finding the derivative of something like $e^{g(x)}$).
A: $$\frac{d (\ln(1-x))}{dx},\quad 0<x<1$$
$$=\frac{d}{dx} \left(-\sum_{k=1}^\infty \frac{x^k}{k}\right)$$
$$=-\sum_{k=1}^\infty x^{k-1}$$
$$=-\frac{1}{1-x}$$
