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Differentiation of e^x and lnx

  1. Nov 8, 2007 #1
    please explain in more detail on how we come up with the answers below. Thanks in advance!
    (formulas much appreciated)

  2. jcsd
  3. Nov 8, 2007 #2


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    There are many websites on the internet explaining those two.

    Just search for the derivative for the first one though. Because for the second one, you can use implicit differentiation as follows...

    y = lnx (use exponential)

    e^y = x (differentiate implicitly)

    y' e^y = 1

    y' = 1/(e^y) (now sub in y)

    y' = 1/(e^(lnx)) = 1/x
  4. Nov 12, 2007 #3
    The first one is found using the rule for differentiating the inverse of a function,

    [tex](f^{-1})^{'}(x) = \frac{1}{f^{'}(f^{-1})} [/tex]

    because the exponential function is the inverse of the natural logarithmic function.


    [tex]\frac{d}{dx}e^{x} = \frac{1}{\frac{d}{dx}[ln (e^{x})]} [/tex]

    [tex]\frac{d}{dx}e^{x} = \frac{1}{\frac{1}{e^{x}}}[/tex]

    [tex]\frac{d}{dx}e^{x} = e^{x}[/tex]
  5. Nov 12, 2007 #4


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    One point I think should be made: watch what you say! In particular, watch where you put "=". The most natural interpretation of
    is that you are saying that y is equal to both ln x and 1/x- so they are equal to each other.

    Be very careful that you say
    y= ln x,
    y'= ln x.

    As to how you prove them, that depends strongly on how you define the functions!
    If you define ax by first defining an= "product of a with itself n times" as long as n is a positive integer, then extend to all rational numbers by requiring that ax+y= axay and axy= (ax)y for x,y any rational numbers, and finally extending to any real number by "continuity", then you can use the "difference quotient":
    [tex]\lim_{h\rightarrow 0}\frac{a^(x+h)-a^x}{h}= \lim_{h\rightarrow 0}\frac{a^xa^h- a^x}{h}= \lim_{h\rightarrow 0}\frac{a^h- 1}{h} a^x[/itex]
    You can prove that [itex]\lim_{h\rightarrow 0}(a^h- 1)/h[/itex] exists for a any positive number, so that the derivative of [itex]a^x[/itex] is a constant times [itex]a^x[/itex] and then define e to be the number such that that constant is 1.
    Having done that, define ln(x) to be the inverse function to [itex]e^x[/itex]. If y= ln(x) then x= [itex]e^y[/itex] so [itex]dx/dy= e^y= x[/itex] and then [itex]dy/dx= 1/x[/itex].

    Conversely, it has become common in calculus books to define ln(x) as [itex]ln(x)= \int_1^x (1/t)dt[/itex] so that we have immediately from the Fundamental Theorem of Calculus that d(ln x)/dx= 1/x. Then define exp(x) to be the inverse function to ln(x) and we have: if y= [itex]e^x[/itex] then x= ln(y) so dx/dy= 1/y and dy/dx= y= [itex]e^x[/sup].
  6. Nov 12, 2007 #5
    I'm afraid that the OP may lose the point you were trying to make if you leave this as it is, otherwise I wouldn't bother correcting something so obvious.

    Also you put parentheses where you meant to put braces.
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