# infinite derivative of $e^x$

i have started thinking about one topic a few days ago and i am confused if i am wrong or what happens,generally we know that function $e^x$ is somehow 'magic',which means that derivative and integral of this function is again $e^x$(let reject constant term during the integral).but on the other hand we can say that (here let's use d as sign of derivative)

$$d(e^x)=d(e*e^{x-1})$$

which is equal to $d(e)*e^{x-1}+ e*d(e^{x-1})$

because of feature of derivative of two function(in this case our function $f(x)=e$ is constant),clearly first term is zero,so we have $e*d(e^{x-1})$,if we continue it to infinite time, we can see that in derivative sign power approaches $x$,or something like this

$$d(e^{x-1}),d(e^{x-2}),d(e^{x-3})$$ and at the same time power of constant $e$ is increasing corresponding,but my confusion is that does never power in

$$d(e^{x-c})$$ where $c$ is some constant is changed from -infinite to +infinity,but does it never make $e^{x-c}$ as a constant? or does never equal $e^{x-c}$ never equal to $1$? meaning that $x=c$? if ti makes constant ,then we know that derivative of constant is zero and whole multiplication becomes zero,which is contradiction what $$d(e^x)=e^x$$

sorry if my idea seems stupid,but i am curious in this topic and please help me to clarify everything

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Instead of indenting, which can prevent latex from rendering, use double dollar signs to center a math formula on its own line :) –  Jim Mar 20 '13 at 6:48
thanks $Jim$ for editing –  dato datuashvili Mar 20 '13 at 6:48

When we write $f(x) = e^x$ the $x$ is a variable, it does not have a prescribed value. So $e^{x-c}$ is not a constant.

It is true that there is a particular value of $x$ for which $e^{x - c} = 1$ because if you pick, for example, $c = 3$ and look at the graph of $e^{x-3}$ it does intersect the horizontal line $y = 1$ (it happens when $x = 3$). But that's only one value of $x$. The graph of $e^{x-c}$ is not a horizontal line so the function $e^{x-c}$ is not constant.

The thing to remember here is that the derivative of a function does not depend on the value of that function at a single point. Instead it depends on how the function behaves around that point. So even though for every $x$ we could pick a $c$ such that $e^{x - c} = 1$, without changing that $c$ it still would not be the case that $e^{x-c}$ is constant around that particular $x$ so it's derivative will still not be $0$.

Also, remember you can't really say $c$ is infinity because infinity is not a real number that you can plug into an equation. You can take limits as values tend towards infinity but that's not the same thing as plugging it in.

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ok thanks @Jim now it makes sense for me –  dato datuashvili Mar 20 '13 at 7:04

Hmmmmmmmmmm well \begin{eqnarray*} d(e^x)&=&d(e\cdot e^{x-1})\\ &=&d(e)\cdot e^{x-1}+e\cdot d(e^{x-1})\\ &=&0\cdot e^{x-1}+e\cdot d(e^{x-1})\\ &=&e\cdot d(e^{x-1})\\ &=&e\cdot (e^{x-1})\\ &=&e^x. \end{eqnarray*}

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we are not stoped @Fixed Point ,i meant when we continue derivation procedure –  dato datuashvili Mar 20 '13 at 7:00
Yep repeat this as many times as you want. It'll work. –  Fixed Point Mar 20 '13 at 7:01