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Derivative and integral of e^anything

  1. Aug 7, 2011 #1
    the integral and derivative of e^x is e^x itself.

    I was told that the derivative and integral of e to the ANYTHING power is e to that something power, meaning that:

    ∫(e^(6x+4x²+5y³))dx is e^(6x+4x²+5y³)


    d/dx(e^(6x+4x²+5y³)) is e^(6x+4x²+5y³)

    However I recently saw an equation in which ∫(e^(ln(2)*x))dx equalled (1/ln(x))*e^(ln(2)*x)

    I do not understand what this is…. They said that this had something to do with the reverse chain rule, but this confused me even more…

    If the exponent is seen as an "inside part of the function", then


    would = (d/dx)*y*x^(y-1)

    instead of just y*x^(y-1)

    I'm very confused whether exponents are "inside functions" or not and why ∫(e^(ln(2)*x))dx did not equal e^(ln(2)*x)

    Did I learn the wrong information? What am I not understanding? Please help D=
  2. jcsd
  3. Aug 7, 2011 #2
    The "ANYTHING" you referred to only includes constants (so it's not really anything). If you have e to the some function of x, you need the chain rule for differentiation.
  4. Aug 7, 2011 #3
    so x is constant, but ln(2)*x is not?

    how do I differentiate between those two situations?

    would 6x+4x²+5y³ be constant or a function of x?
  5. Aug 7, 2011 #4
    Sorry, I was somehow a bit confused. The only function that equals to its derivative is [itex]e^x[/itex]. You have to use the chain rule for other cases.
  6. Aug 7, 2011 #5
    ohhhh. so derivative of e^2x would be 2*e^2x?

    and integral of e^2x would be (e^2x)/2?
  7. Aug 8, 2011 #6
    Yupp :)
  8. Aug 8, 2011 #7
    so how would you integrate ex^2?
    Last edited: Aug 8, 2011
  9. Aug 8, 2011 #8


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    By the chain rule, the derivative of [itex]e^{f(x)}[/itex]
    is [itex]e^{f(x)}f'(x)[/itex]
    so that, for example,the derivative of [itex]e^{2x}[/itex] is [itex]e^{2x}[/itex] times the derivative of 2x which is 2- the derivative of [itex]e^{2x}[/itex] is [itex]2e^{2x}[/itex]. The derivative of [itex]e^{x^2}[/itex] is [itex]e^{x^2}[/itex] times the derivative of [itex]x^2[/itex] which is 2x. The derivative of [itex]e^{x^2}[/itex] is [itex]2xe^{x^2}[/itex].

    However, since the integral is defined "inversely" (the integral of f(x) is the function whose derivative is f(x)), there are no general "rules" for integration and most elementary functions cannot be integrated in terms of elementary functions. In particular, while [itex]e^{x^2}[/itex] is continuous and so has an integral, it cannot be written in terms of elementary functions.

    It can be written in terms of the "error function", Erf(x). To integrate [itex]e^{x^2}[/itex], let x= iy so that dx= i dy amd [itex]x^2= -y^2[/itex]. Then
    [tex]\int e^{x^2}dx= \int e^{-y^2}i dy= i\int e^{-y^2}dy= iErf(y)= i Erf(-ix)[/tex]
    where "Erf(x)" is defined as the integral of [itex]e^{-x^2}[/itex].
  10. Aug 8, 2011 #9
    This rule also applies to exponents of X or Y right, not just e?

    so the derivative of x^(x²) would be 2x*(x²)*x^(x²-1)?
  11. Aug 8, 2011 #10
    To solve that one, let y = xx2 so that ln(y) = x2ln(x).
    Differentiating both sides yields y'/y = x2/x + 2xln(x) = x + 2xln(x), where y' = dy/dx. Then y' = xx2(x + (2x)ln(x)).
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