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$$\frac{d}{dx}{a^{x}=a^{x}\ln(a)}\tag{1}$$

$$\int{a^{x}dx=\frac{1}{\ln(a)}a^{x}},a>0,a\neq1\tag{2}$$

This is what I found written in my book. I was wondering if they wrote it wrongly:

$$\frac{d}{dx}{a^{x}=a^{x}\ln|a|}\tag{1'}$$

$$\int{a^{x}dx=\frac{1}{\ln|a|}a^{x}},|a|\neq1\tag{2'}$$

Can $(1)$ & $(2)$ be written as $(1')$ and $(2')$. Are $(1')$ and $(2')$ mathematically correct? This (see 'Addendum') might help you in answering the question.

PS: Notice that in (1') and (2'), I did not put absolute value signs around $a$, like $|a|^{x}$, in LHS.

tryingtobeastoic
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2 Answers2

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The function $f(x) = a^x$ is only continuously defined on the entire set of real numbers for positive values of $a$; if $a$ is not positive, the function is not differentiable at all. So any formula for $\frac{d}{dx}a^x$ implicitly assumes $a > 0$ as a condition, which means that the absolute value symbols are unnecessary.

mweiss
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    Maybe include some parts where $f$ uses another function $g$, not the constant $a$ for the absolute value to be explained? – soupless Jan 20 '22 at 03:25
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    Sir, did you notice that in (1') and (2'), I did not put absolute value signs around $a$ in LHS? – tryingtobeastoic Jan 20 '22 at 03:30
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    Yes, my point is that the expression $\frac{d}{dx} a^x$ is only meaningful if we assume $a>0$, in which case there is no difference between writing $\ln |a|$ and $\ln a$. – mweiss Jan 20 '22 at 04:19
  • @soupless I’m not sure I understand what you mean. – mweiss Jan 20 '22 at 04:19
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    @mweiss What I mean is that, maybe you can explain what happens when $f$ is an exponential function of the form $(g(x))^x$? Something like that to explain the absolute value? – soupless Jan 20 '22 at 05:10
  • @tryingtobeastoic Notice that $\displaystyle(-7)^\frac5m$ is undefined whenever $m$ is even. – ryang Jan 20 '22 at 07:54
  • @soupless That would not justify using absolute values. Once again, you are missing the point of the answer. If $g(x)\lt0,$ then the expression $g(x)^x$ is meaningless to begin with, – Angel Jan 20 '22 at 16:27
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Before answering your question, I first need to establish what the notation $a^x$ even means: otherwise, you are likely to just stay confused. The notation $a^x$ is an abbreviation for the somewhat more tedious expression $\exp[\ln(a)x],$ where $\exp$ is a function $\mathbb{R}\rightarrow(0,\infty)$ defined by the functional equation $\exp(x+y)=\exp(x)\exp(y)$ with $\exp(0)=1$ and $\exp(1)=\lim_{n\to\infty}\left(1+\frac1{n}\right)^n=:e,$ and $\ln$ is the inverse function. Typically, we use the notation $\exp(x)=e^x$ when doing calculus, but the definition for $a^x$ stays the same.

Keeping this in mind, you can use the chain rule to conclude that $$\frac{\mathrm{d}}{\mathrm{d}t}a^t=\frac{\mathrm{d}}{\mathrm{d}t}e^{\ln(a)t}=\ln(a)e^{\ln(a)t}=\ln(a)a^t.$$ However, in order to be able to make this conclusion, $\ln(a)$ must be a well-defined expression to begin with, and this is only true if $a\gt0.$ The expression $a^x$ is meaningless when $a\lt0,$ unless you confine $x$ to the set of integers, which is clearly not what we are doing.

I understand this is confusing, which is why I am a strong advocate for stopping the use of such bad notation and making it obsolete. I think we should stick to using the notation $\exp[\ln(a)t],$ even if it takes slightly more symbols to write. Distinguishing this notation from the power notation $x^n,$ where $n\in\mathbb{Z},$ is what I consider a healthy mathematical habit that eliminates confusion and is much less error-prone. \

Anyhow, the fact is that $$\frac{\mathrm{d}}{\mathrm{d}t}a^t=\ln(a)a^t$$ implies that $$\frac{\mathrm{d}}{\mathrm{d}t}\frac1{\ln(a)}a^t=\frac1{\ln(a)}\ln(a)a^t,$$ but it only implies that if $$\frac1{\ln(a)}$$ is a meaningful expression, and such an expression is only meaningful when $a\gt0,\,a\neq1.$ For $a=1,$ we have that $1^t=1,$ so $$\frac{\mathrm{d}}{\mathrm{d}t}1^t=0=0\cdot1^t,$$ and $$\frac1{\ln(0)}\neq0,$$ and furthermore, $$\lim_{a\to1}\frac1{\ln(a)}\neq0.$$ This is why $a\neq1$ is a necessary restriction.

Angel
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