- #1
Some_dude91
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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:
[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]
[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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