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E in Calculus

  1. Sep 28, 2008 #1


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    What if the proof for differentiating and integrating e^x.

    For d/dx (e^x) i used the chin rule and did
    u = x
    y = e^u
    dy/du = ue^u-1
    du/dx = 1
    dy/du*du/dx = ue^u-1
    so you get xe^x-1 but thats not right. I don't even know where to start with intergration.

    Can any one show me? Thanks:biggrin:

    (also where are all the symbols have gone? I Don't know how to put them in anymore.)

    *edit* Sorry I didn't realise this wasn't home work help section. Can a moderator move it there please.
  2. jcsd
  3. Sep 28, 2008 #2


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    Why do you have dy/du=ue^(u-1)?
  4. Sep 28, 2008 #3
    well, there are different ways as how the derivative of e^x is defined. We usually start by the def. of the derivative.

    let [tex]f(x)=a^x[/tex], be any exponential function, so

    [tex](a^x)'=\lim_{h\rightarrow 0}\frac{a^{x+h}-a^x}{h}=\lim_{h\rightarrow 0}\frac{a^x(a^h-1)}{h}=a^x\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex]

    Now, notice that [tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}=f'(0)[/tex] , right? In other words, this is the derivative of the exponential function at the point x=0.

    SO, it looks like the derivative of any exponential function at any point, say c, [tex]f(x)=a^x[/tex] is simply a constant multiple of its value at that particular point. IN other words, the derivative of any exponential function, and the value of the function at a certian point x are proportional.

    Now, it is convinient for us to find an exponential function such that the derivative of that function at x=0 is 1. f'(0)=1.

    [tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}[/tex] let [tex]a^h-1=t=>a^h=t+1[/tex] also

    [tex] h=log_a=log_a(t+1)[/tex] notice that when h-->0, t-->0, so;

    [tex]\lim_{h\rightarrow 0}\frac{a^h-1}{h}=\lim_{t\rightarrow 0}\frac{t}{log_a(t+1)}=\lim_{t\rightarrow 0}\frac{1}{log_a(t+1)^{\frac{1}{t}}}=\frac{1}{log_ae}=lna[/tex]

    Now as we can see, in order for f'(0)=1, a=e, so [tex](e^x)'=e^x[/tex]
  5. Sep 28, 2008 #4


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    because u is a value. the power is decreased by one.

    Thanks for the help.
  6. Sep 28, 2008 #5


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    [tex]\frac{d}{du} u^e =eu^{e-1}[/tex]


    [tex]\frac{d}{du} e^u=ue^{u-1}[/tex]

    To find the derivative of e^x, you will need to use the definition of derivative in terms of limits.
  7. Sep 28, 2008 #6
    The power rule of differentiating does not apply here, since your power here is not a number, but rather a function/variable.
    Like i said in my above post. In my previous post, i explained how why the derivative of exp. functions is what it is.

    other ways of doing it, which come indirectly from the def. of derivatives, is

    let [tex]f(x)=a^x[/tex] then let [tex] a^x=y=>lny=xlna[/tex] now let's differentiate implicitly:

    [tex] \frac{y'}{y}=lna=>y'=ylna=>y'=(a^x)'=a^xlna[/tex] in your case you have [tex] f(x)=e^x[/tex] so all you need to do is replace a by e. and youll get what u need.!
  8. Sep 28, 2008 #7


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    I have absolutely no idea what you mean by "u is a value"! Every symbol or expression in mathematics is a value! The question is whether or not u is a variable or a constant. Since you let u= x and x is a variable, so is x.

    The derivative of xa, where a is a constant, with respect to x, is a xa-1 but the derivative of ax is ln(a) ax and in the special case that a= 2, ln(e)= 1, that becomes d(ex)/dx= ex.
  9. Sep 29, 2008 #8


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    Well my working was wrong. I just made u = x (like in the chain rule, e^2x-1 u = 2x-1) and I took 1 off it because when you differentiate the power decreases by 1.
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