Is there a way to prove derivative of e^x with such conditions: 1)Without using implicit differantiation. 2)Using compound interest definition (or using any other definition but with proof that compound interest definition and other definition are equal). 3)Without using fact that derivative of ln x is 1/x (or using this fact but with proof that derivative of ln x is 1/x).
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1Can I ask why you desire such an answer? – Tom Himler Jan 09 '18 at 19:13
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5How do you define the exponential function? – NickD Jan 09 '18 at 19:14
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The answer depends on how you define the exponential. – copper.hat Jan 09 '18 at 19:34
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@Nick As number e to the power of x. And how I define number e you can find in the main question. – Юрій Ярош Jan 09 '18 at 20:20
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See this answer: https://math.stackexchange.com/a/541330/72031 – Paramanand Singh Jan 10 '18 at 04:03
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write $$\frac{e^{x+h}-e^x}{h}=\frac{e^x(e^h-1)}{h}$$ and compute the Limit if $h$ tends to Zero to prove the Limit set $$e^h-1=t$$ then we get the term $$\frac{1}{\ln(1+t)^{1/t}}$$
Dr. Sonnhard Graubner
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You can use the fundamental theorem of calculus: $$\left \lvert \frac{ e^h - 1 }h - 1 \right \rvert = \frac{1}{h} \left \lvert \int^h_0 (e^t - 1) dt \right \rvert \le \frac{1}{h} \int^h_0 \lvert e^h - 1 \rvert dt = \lvert e^h - 1 \rvert \to 0$$ and so $\frac{ e^h - 1 }h \to 1$. – User8128 Jan 09 '18 at 19:38
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my proof is better, i don't use this Theorem, since we want to prove this Theorem without using calculus – Dr. Sonnhard Graubner Jan 09 '18 at 19:40
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How about $e^x=\sum_{k=0}^\infty\frac{x^k}{k!}$?
Pink Panther
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No, you can't. It's just a taylor series of e^n and the existence of taylor series depends on the derivative of e^x. You need to prove it in another way. – Xwtek Feb 10 '21 at 11:09
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depending on where you're looking that can be a property or a definition... – Pink Panther Feb 10 '21 at 19:04
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Easiest way is using a power series $$ f(x) = e^x = \sum_{k=0}^\infty x^k/k! $$ and so $$ f'(x) = \sum_{k=1}^\infty \frac{x^{k-1}}{(k-1)!} = \sum_{n=0}^\infty x^n/n! = e^x. $$
gt6989b
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Basically, you have to prove the limit, $$\lim\limits_{x\to 0}\frac{e^x-1}{x}=1.$$
Now $$\left(1+\frac{x}{n}\right)^{n+1}$$ is decreasing for $x\leq 1$ and thus you have for all $n$,
$$1\leq \frac{e^x-1}{x} \leq \frac{1}{x}\left[\left(1+\frac{x}{n}\right)^{n+1}-1\right]$$ and the limit on the left as $x\to 0$ is $\frac{n+1}{n}$. From this the limit follows rigorously.
Rene Schipperus
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