I am posting this question, in order to make something clear, since i am confused by derivation of the exponential function. I'll post the formula i used, correct me if you find something wrong thank you:(adsbygoogle = window.adsbygoogle || []).push({});

[itex]

{\frac{d}{Dx}}\ e^x =>

[/itex]

[itex]

{\frac{e^{x + Dx} - e^x}{Dx}}\ =>

[/itex]

Here i factored out [itex] e^x [/itex]

[itex]

e^x\ {\frac{e^{Dx} - 1}{Dx}}\ =>

[/itex]

I integrated the exponential

[itex] e^{Dx}\ =\ cosh(Dx)\ +\ sinh(Dx)\ [/itex]

Now based on theory

[itex] e^{Dx} [/itex]

already approaches a value which is a tiny little above 1 so if we subtract one it will be one tiny little value above 0 and if we divide by [itex] Dx [/itex]

we are likely to get 1 as the values approach each other as they shrink to lower values.

But as i try to factor for [itex] sinh(x) [/itex] i take that since

[itex]

coth(x)\ =\ {\frac{\cosh(x)}{sinh(x)}}\

[/itex]

i take that

[itex] cosh(Dx)\ =\ coth(Dx)\ *\ sinh(Dx) [/itex]

[itex]

e^x\ {\frac{coth(Dx)sinh(Dx) + sinh(Dx) - 1}{Dx}}\ =>

[/itex]

[itex]

e^x\ {\frac{sinh(Dx)\ (coth(Dx) + 1)}{Dx}}\ -\ {\frac{1}{Dx}}\

[/itex]

As sinh(Dx) approaches 0 so does Dx and they tend to be 1 in division

[itex]

e^x\ *\ (coth(Dx)\ +\ 1)\ -\ {\frac{1}{Dx}}\

[/itex]

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# Question about calculus, on derivatives

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