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Another way. $\exp(-z^2)$ is an entire function. Its power series at the origin converges for all $z$. Take the anti-derivative of that series term-by-term. It still converges for all $z$. (Use the formula for radius of convergence.) So the series of antiderivatives is an antiderivative for $\exp(-z^2)$. That argument works for all entire functions. ...

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The answer to your question depends on what you expect your readers to know in advance. Since you're writing about derivatives, they should know calculus. Then the fact that $e^x$ is its own derivative is likely to be a better known prerequisite than the statement about logarithmic derivatives you want to use to prove it.

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You haven't really proven anything. You have used something that needs proving, to "prove" that the derivative of $\mathrm e^x$ is itself. It's really just a circular argument. I would suggest that you prove it from first principles: $$\mathrm f'(x) = \lim_{h \to 0} \left(\frac{\mathrm f(x+h)-\mathrm f(x)}{h}\right)$$ In your case, $\mathrm f(x) = \mathrm ... 3 Define, for some$z_0\in \mathbb{C}$$$g(z)=\int_{z_0}^z e^{-t^2}dt$$ Now for any$2$homotopic (within the domain where$e^{-z^2}$is analytic) curves the value of this integral will be the same. Since$e^{-z^2}$is analytic and$\mathbb{C}$is simply connected, we find that the value of$g(z)$is independent of the chosen contours. Hence this is a well ... 3 There is a mistake in your algebra. From your third equation $$\int\frac{\log_a(e)}{x}\,dx=\log_a(x)+C\tag1$$ you can deduce that for any$c$$$\int\frac cx\,dx={c\over\log_a(e)}\log_a(x)+C.\tag2$$ (This is equivalent to the Wolfram result). Comparing (2) to your second equation, it doesn't follow that$c=\log_a(e)$. 2 If$M$is complete, then$D_1$is open, which follows essentially from the fact that$p \mapsto r_p$is continuous. For instance, consider the map$f : TM \to TM$given by $$f : v \mapsto \frac{v}{r_p},$$ where$p$is the footpoint of$v$;$f$is continuous, and your$D_1$is the inverse image under$fof the open set $$\lbrace v \in TM : ||v|| < ... 2 It is the natural log! Think about it this way. Let y=\ln(x). It follows that e^y=x. From there, we can differentiate implicity; y'*e^y=1. Now, e^y=x, so x*y'=1, and y'=1/x. 2 The base is the one that satisfies$$\begin{align*} \frac1x &= \frac{d}{dx}\log_a x\\ &=\lim_{h\to0}\frac{\log_a(x+h)-\log_a x}{h}\\ &= \lim_{h\to0}\log_a\left(1+\frac hx\right)^{1/h}\\ &= \lim_{h\to0}\log_a\left(1+\frac hx\right)^{x/hx}\\ &= \frac1x\log_a\left[\lim_{h\to0}\left(1+\frac hx\right)^{x/h}\right]\\ \end{align*}$$It ... 2 The composition of functions with an antiderivative has an antiderivative. \exp has an antiderivative because it is its own derivative (and hence its own antiderivative) everywhere. z \mapsto -z has an antiderivative z \mapsto \frac{-z^2}{2}. z \mapsto z^2 has antiderivative z \mapsto \frac{z^3}{3}. 1 For the record, I agree with @EthanBolker's answer above. With regards to the question of whether a proof of the fact exists without assuming the derivative of e^x is e^x, the answer is yes. https://en.wikipedia.org/wiki/Logarithmic_derivative https://proofwiki.org/wiki/Derivative_of_Natural_Logarithm_Function The equality follows from the chain rule, ... 1 Because it is exponential decay, I would prefer to use A(t)=A(0)e^{-kt}. Then \frac{300}{375}=e^{-72k}, so k=\frac{1}{72}\ln(375/300). Now for the half-life \tau, we have \frac{1}{2}A(0)=A(0)e^{-k\tau}, so k\tau=\ln 2. It follows that the half-life, in hours, is 72\frac{\ln 2}{\ln(375/300)}. 1 Set s^{2\alpha+1}=w. Then$$ I = \int_0^t s^{2\alpha - 1} \exp\left(\frac{i \sqrt{2} \left(t^{2 \alpha + 1} - s^{2 \alpha + 1}\right)}{ 2 \alpha + 1}\right)\mbox{d}{s}=(1+2\alpha)^{-1}\exp\left(\frac{i\sqrt{2}t^{2\alpha+1}}{2\alpha+1}\right)\int_0^{t^{2\alpha+1}}\frac{dw}{w^{\frac{1}{2\alpha+1}}} \exp\left(\frac{-i\sqrt{2}w}{2\alpha+1}\right)  ... 1 Repeating the quotation from Wikipedia: However, logarithms in other bases differ only by a constant multiplier from the natural logarithm, and are usually defined in terms of the latter. There is nothing here that says this property of the natural logarithm is unique. Indeed, if \begin{align} \log_a(x) &= A\log_e(x) &&\text{and}& ... 1 You're just misunderstanding what the article (in Wikipedia, here) says. The quote that explains why the natural log is uniquely called "natural" is this one: The natural logarithm can be defined for any positive real numbera$as the area under the curve y =$1/x$from$1$to$a$(the area being taken as negative when$a<1$). The simplicity of this ... 1 Hint. Assume$x>0$. Then you get $$f'(x)=\frac{-2x\times (1+2 \ln x)}{x^{2 x^2}} .$$ Can you take it from here? 1 By definition, $$f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}.$$ For$a > 0$and$f(t) = a^t$, we get $$f'(t) = \lim_{h \to 0} \frac{a^{t+h} - a^t}{h}.$$ Now let$t = 0$: $$f'(0) = \lim_{h \to 0} \frac{a^{0+h} - a^0}{h} = \lim_{h \to 0} \frac{a^h - 1}{h}.$$ 1 The limit is the same as the one that defines$f'(0)$using the definition of derivative applied to this function$f$. The harder part of the proof, not written in the question, is to show that this limit exists and to determine its value (a logarithm) as a function of the base$a$. The calculation posted in the question is the easier algebraic step of ... 1 HINT: Enforce the substitution$s\to s^{-1/b}\$ to yield the integral $$\frac1b \int_{t^{-1/b}}^\infty e^{-\frac12 a^2s^s}\,ds$$

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