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Homework Help: Having Trouble With The Chain Rule

  1. Jun 13, 2006 #1
    [tex]f(t) = 6e^{0.013t} [/tex]

    How do I find the derivative of this?

    I'm confused. Do I have to use the chain rule here or the product rule, or both?
  2. jcsd
  3. Jun 13, 2006 #2
    You will have to use the chain rule on this. Do you know the derivative of e^x?

    Example of the chain rule.
    [tex]f(x) = 5(2x+1)^3[/tex]. To find the derivative, use the chain rule.

    [tex] dy/dx = 5 ( 3(2x +1)^2 *2). = 30(2x+1)^2 [/tex].

    If it helps you, you can call 2x +1 "u" and say [tex] dy/dx = dy/du * du/dx [/tex].

    du/dx = 2. dy/du = 3u^2. Don't forget the constant five outside.
  4. Jun 13, 2006 #3
    Isen't the derivative of e^x itself? or do I need to use the natural log.

    Here's my attempt from looking at your example :

    [tex] 6(0.013t(e)^{0.013t} * e^{0.013t} [/tex]
    Last edited: Jun 13, 2006
  5. Jun 13, 2006 #4

    The derivative of [tex] e^u = e^u * du/dt [/tex].

    So, Let's call u = .013 t

    Now, we have [tex] e^u [/tex] We know the derivative of [tex] e^u = e^u du/dt [/tex].

    So, [tex] dy/dt = e^u * du/dt. [/tex]

    What is du/dt? We know u = .013t. Can we get du/dt from that? Does this help a bit?
  6. Jun 13, 2006 #5
    Hmm, i'm not sure if I'm getting the whole U thing sorry.

    I have the chain rule being D/DX F(G(X)) = F'(G(X)) * G'(X)

    I'm not even sure which part of the equation is F and which is G. If someone could explain that to me, I might have a better chance of plugging this in right.

    I think it's the exponents, and the e that is throwing me off.
    Last edited: Jun 13, 2006
  7. Jun 13, 2006 #6
    I think you're being confused by the notation. I'll attempt to clear it up, then I"ll try to help you out.
    F(g(x) means we have a function inside another function. So, let's look at what we have here:

    e^.013t. That .013t is contained in an "e^x" like function, where x = .013t. Well, we can call that .013t = G(t) if we want.

    Now, we want an F(g(t). We're going to call e^x "F(x)." Now, we have F(x) = e^x. . What does this mean?

    Well, it means we can replace x for something like F(1) = e^1. Or, F(5) = e^5.

    Or, usefully:

    F(g(t)) = e^(g(t)) = e^.013t

    So, Now we have that F(g(t)) form. The differentiation says

    F'(g(t)) * g'(t).

    This means we need two things: We need the derivative of F'(g(t)) (the derivative of e^x) and the derivative of G(t) (.013t) and then we multiply them together.

    Does this help?


    y = (x^2 + 3)^3 .

    We want to express it in the form of F(g(x)). Notice how we have an inside function (x^2 + 3) and an outside function a Something^3. Well, we'll call the inside function g(x) = x^2 + 3. That outside ^3 function will be called F(u) = u^3

    So, we have [tex]F(g(x)) = (x^2 + 3)^3 [ /tex]

    The chain rule says "Take the derivative of the outside funciton and multiply it by the derivative of the inside funciton" or F'(g(x)) * g'(x)

    So, the derivative of the outside function ^3 = 3u^2. (Remember, that "u" here is actually standing for our x^2 + 3)
    The derivative of our inside function (the x^2 +3 ) is G'(x) = 2x

    Now, Multiply 3u^2 * 2x

    Remember, we said u = g(x) = x^2 +3

    So, our derivative is

    3(x^2 +3)^2 *2x

    The trick here is to see an "Inside function" and an "Outside function". THe outside function contains the inside function. Does this help at all, or do you want me to help walk you through it a bit more?
  8. Jun 14, 2006 #7
    Thanks for that write up.

    I tried again and I got this: [tex] f'(t) = .013*6e^{-.987T}(.013)[/tex]

    I felt like I was supposed to do something with the 6 but wasn't sure.
  9. Jun 14, 2006 #8

    matt grime

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    Science Advisor
    Homework Helper

    the derivative of e^x is e^x, it is not e^{x-1}.
  10. Jun 14, 2006 #9
    Here, perhaps an example of an actual exponential function would help.

    f(x) = ce^(ax^2)

    Where c and a are constants.

    Let u = ax^2

    Let f(u) = e^u

    So, the derivative dy/dx

    [tex] dy/dx = dy/du * du/dx [/tex]

    8So, We can find du/dx = 2ax.

    We can find dy/du = e^u. (The derivative of e^x is e^x, as Matt pointed out).

    So, fill in where things go, dy/du = e^u du/dx = 2ax

    [tex]dy/dx = 2cax * e^u [/tex] Remember how I said u = ax^2? So replace u.

    dy/dx = 2cax e^(ax^2)

    Another example

    8e^9x = y.

    dy/dx = dy/du * du/dx.
    u = 9x

    du/dx = 9

    8e^u = y. Dy/du = e^u

    8e^u *9 = dy/dx = 72e^u = 72e^9x

    Do these examples clear it up a bit more?
  11. Jun 14, 2006 #10
    Yes, thanks for your time and effort.

    Now I get :

    [tex]f(t) = 6e^{0.013t}[/tex]

    [tex]let u = 0.013t[/tex]

    [tex]f(t) = 6e^u[/tex]

    [tex]\frac {dy}{dt} = \frac {dy}{du} * \frac {du}{dt}[/tex]

    [tex] 6e^{0.013t} * (0.013) [/tex]
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