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!
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1$\begingroup$ Could you give an example of a formula you're struggling to remember? $\endgroup$– aqualubixFeb 3 at 15:21
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3$\begingroup$ Have you tried coming up with the “formulas” yourself? That always helped me $\endgroup$– bananapeel22Feb 3 at 15:26
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1$\begingroup$ E.g. $f(x)=a^x = e^{x\ln(a)}$ and so by using the rules of differentiating $e$ we get $f’(x)=\ln(a)e^{x\ln(a)}=\ln(a)a^x$ $\endgroup$– bananapeel22Feb 3 at 15:33
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$\begingroup$ @aqualubix I can't remember log_a{x}' = 1/ (xlna) $\endgroup$– Eternal BladeFeb 11 at 14:43
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$\begingroup$ @bananapeel yea but I don't usually do that because there isn't much time in an exam $\endgroup$– Eternal BladeFeb 11 at 14:43
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3 Answers
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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: $$\frac{\mathrm d}{\mathrm dx} x^{x}=\frac{\mathrm d}{\mathrm dx} \left(e^{x\ln x}\right)=\left(x\left(\frac1x\right)+\ln x\right)e^{x\ln x}=(1+\ln x)x^x.$$
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$\begingroup$ Just wondering, is there a reason you use $:=$ to define $a^x$, but not to define $\log_a(x)$? Is the latter not a definition? $\endgroup$ Feb 3 at 23:19
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1$\begingroup$ @user170231 Yes, the latter is derived from $(y=b^x{\overset{\text{def}}{\iff}}x=\log_by)$ for $b\in(0,1)\cup(1,\infty).$ I might as well add that the above definition has the qualification $a>0.$ $\endgroup$– ryangFeb 3 at 23:41
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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 $e^{g(x)}$).
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1$\begingroup$ exactly what I used to do when I used to forget $d/dx \ln x$ (+1) $\endgroup$– D SFeb 3 at 15:30
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$$\frac{d (\ln(1-x))}{dx},\quad 0<x<1$$
$$=\frac{d}{dx} \left(-\sum_{k=1}^\infty \frac{x^k}{k}\right)$$
$$=-\sum_{k=1}^\infty x^{k-1}$$
$$=-\frac{1}{1-x}$$
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2$\begingroup$ Why use power series when this is a very simple chain rule problem? Surely this won't help the OP with their memorization. $\endgroup$ Feb 3 at 15:44
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$\begingroup$ @EthanBolker; its pretty simple to use power series too! $\endgroup$– JMPFeb 3 at 15:48
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4$\begingroup$ Yes, but not if you're just learning calculus ... $\endgroup$ Feb 3 at 16:20