I was searching for the proof of [tex]\frac{d}{dx} e^x = e^x[/tex].(adsbygoogle = window.adsbygoogle || []).push({});

and I found one in yahoo knowledge saying that

[tex]\frac{d}{dx} e^x = \lim_{Δx\to 0} \frac {e^x(e^{Δx}-1)} {Δx}[/tex]

[tex] = \lim_{Δx\to 0} \frac {e^x [\lim_{n\to\infty} (1+ \frac{1}{n})^{n(Δx)}-1]} {Δx}[/tex]

Let [tex] h= \frac {1}{n} [/tex], So that [tex] n = \frac {1}{h} [/tex]

That makes

[tex] \lim_{Δx\to 0} \frac {e^x * [\lim_{h\to 0} (1+ h)^{\frac {1}{h}(Δx)}-1]} {Δx}[/tex]

And it said variable h of the power [tex]\frac{1}{h}[/tex] and that of the limit [tex] \lim_{h\to 0}[/tex] of [tex]e^{Δx} = \lim_{h\to 0} (1+ h)^{\frac {1}{h}} [/tex] can be replaced by [tex]Δx[/tex], to make the whole formula becomes

[tex] = \lim_{Δx\to 0} \frac {e^x * [\lim_{Δx\to 0} (1+ Δx)^{\frac {1}{Δx} * {Δx}}-1]} {Δx}[/tex]

because both variables are going to zero and that they can "replace" each other.

Of course, the calculation can still come to the answer that [tex]\frac{d}{dx} e^x = e^x[/tex].

But I wanna ask if it is true that, two different variables(neither of it is a function of another) approaching to the same value can be "replaced" by each other, or in other words, equal?

If that's true, what are the situations that this kind of "replacements" can't be done.

Also, if that's not true,is there any other way to prove that the derivatives of e^x =e^x without the use of any logarithms?

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# Proofing the derivatives of e^x from the limit approach

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