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Derivative of an Exponential Function

  1. Jan 29, 2016 #1

    Drakkith

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    1. The problem statement, all variables and given/known data
    Find the first and second derivative of the following function:
    F(x)=e4ex

    2. Relevant equations
    d/dx ex = ex
    d/dx ax = axln(a)

    3. The attempt at a solution
    I know the derivative of ex is just ex, but I'm not sure how to go about starting this one. I'm near certain I need to use the chain rule, but I have no idea how to go about starting it. I even tried to substitute u= e4e, but realized that I'm not raising everything to the power of x, only the top e.

    Any ideas?
     
  2. jcsd
  3. Jan 29, 2016 #2

    andrewkirk

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    $$F(x)=e^{4e^x}$$
    Define ##g(u)=4e^u## and ##f(v)=e^v##.
    Then ##F(x)=f\bigg(g(x)\bigg)=f\circ g(x)##, and we can apply the chain rule to ##f## and ##g##.
    $$\frac{d}{dx}\bigg(f\circ g(x)\bigg)=\frac{df}{dg(x)}(g(x))\cdot\frac{dg}{dx}(x)$$
     
  4. Jan 29, 2016 #3

    Drakkith

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    Oh, so you can do a u-sub after all.
    If u=4ex, then d/du eu = eudu/dx
    Then: e4ex(4ex)d/dx
    Finally: 4ex(e4ex)
    Look good?
     
  5. Jan 29, 2016 #4

    SteamKing

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    Use the Law of Exponents to simplify your final result above.
     
  6. Jan 29, 2016 #5

    Student100

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    Yeah that's good, you can simplify it more if you'd like.
     
  7. Jan 29, 2016 #6

    Drakkith

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    Roger.
    4exe4ex = 4ex+4ex

    2nd derivative:
    4ex+4ex
    Let u = x+4ex
    d/du4eu = 4eudu/dx
    4eudu/dx = 4eu(1+4ex)
    Finally: 4ex+4ex(1+4ex)
     
  8. Jan 29, 2016 #7

    Drakkith

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    What's a good way to think about the chain rule when you have exponents raised to exponents? Usually the students at my school are told to think of the chain rule as an inner and outer function. Unless I'm missing something obvious, that seems a bit difficult here.
     
  9. Jan 29, 2016 #8

    andrewkirk

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    The trouble with the terminology of 'inner' and 'outer' functions is that it's notation-dependent. It assumes the function composition is written as f(g(x)). As you've noted, it doesn't work like that for exponents.

    I find it easier to think in terms of order of evaluation. That is, if we were given a numeric value of x, and had to evaluate the function value on a calculator, which function would we do first, which second, and so on if necessary. There is only one way that can be done, so it's not notation-dependent. We can call the functions first, second, third etc, rather than inner and outer.

    If evaluating ##e^{4e^x}## for a given numeric value of x, we'd have to evaluate the expression in the exponent first, so that is the first function. We then need to exponentiate the result of that, and that exponentiation is the second function.

    This approach mirrors how things are written if we use function composition notation, like ##(f\circ g)(x)##. In that notation, we work from right to left in applying the functions, so the rightmost function is first, the next-rightmost second, and so on.
     
  10. Jan 29, 2016 #9

    Drakkith

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    Thanks, andrewkirk. That makes sense.
     
  11. Jan 30, 2016 #10

    ChrisVer

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    Another way:
    [itex] F(x)= e^{4e^x}[/itex]
    [itex] \ln F = 4e^x[/itex]
    [itex] \frac{d}{dx} \ln F = 4e^{x}[/itex]
    it's easier to do the chain rule on the function of ln; because you immediately see that in a function of ln you have another function F, so you clearly have something like G( F(x) ):
    [itex] \frac{d}{dx} \ln F= \frac{1}{F} F' = 4 e^x \Rightarrow F'= 4e^x F[/itex]
    if you want you can redo a derivative on [itex]\frac{1}{F} F' = 4e^x[/itex] and get:
    [itex] \frac{d}{dx} \Big( \frac{1}{F} F' \Big) = \frac{F'' \cdot F - (F')^2}{F^2} = 4 e^x[/itex]
    *

    So:
    [itex]F'' = \frac{4 e^x F^2 + (F')^2}{F}= \frac{4 e^x F^2 + 16 e^{2x} F^2}{F} = 4 ( 1 + 4 e^{x} ) e^x e^{4e^x}[/itex]


    *: If you wanted to avoid the 1/F 's you could take the derivative of :
    [itex] F' = 4e^x F [/itex], giving:
    [itex]F''= 4e^x F + 4 e^x F' = 4e^x F + 4 e^x (4 e^x F) = 4 e^{x} (1+4 e^x) F[/itex]
    just replace F and you get the same result.
     
  12. Jan 30, 2016 #11

    PeroK

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    It may not help, but here's how I think about the chain rule:

    1) Note that for any differentiable function ##f##, there is a well-defined function ##f'##.

    2) If we have ##h(x) = f(g(x)) = (f \circ g)(x)##. Then:

    ##h'(x) = f'(g(x))g'(x)##

    I would read this as: "The function ##h'## evaluated at ##x## equals the function ##f'## evaluated at ##g(x)## times the function ##g'## evaluated at ##x##".

    In your example, I would have taken:

    ##f(x) = e^{4x}## and ##g(x) = e^x##
     
  13. Jan 30, 2016 #12

    SammyS

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    It may simply be a matter of notation.

    The exponential function is sometimes written as exp(x) which is more in accordance with usual function notation.

    Of course, this is defined as exp(x) = ex

    Then the function you ask about in the OP is:

    F(x) = exp( 4*exp(x) ), which looks more like the composition of two functions.

    Thus it may be easier to see how you apply the chain rule.
     
  14. Jan 30, 2016 #13

    Drakkith

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    What do you mean? Is the base of the exponential function always e when defined this way?
     
  15. Jan 30, 2016 #14

    SammyS

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    Yes.
    Of course, you could do something similar for other bases.

    However, when I need to differentiate or integrate an exponential function with some other base, I usually change it to an exponential with base e.

    ##\displaystyle a=e^{\ln(a)} \ ##, therefore, ##\displaystyle a^x=\left(e^{\ln(a)}\right)^x=e^{\ln(a)x} \ ##

    Thus, ##\displaystyle a^x=\text{exp}(\ln(a)x) \ ##.
     
  16. Jan 30, 2016 #15

    Drakkith

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    Alright. Thanks Sammy!
     
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