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Math Help - Chain Rule clarification

  1. #1
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    Exclamation Chain Rule clarification

    I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.

    I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.

    However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?

    If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?

    Please correct anything I am mistaken on. Thank you.
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  2. #2
    Senior Member Spec's Avatar
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    Your reasoning is correct. Try the substitution u=2x to make it even clearer.

    \frac{d}{dx}\left(e^u\right)=\frac{d}{du}(e^u) \cdot \frac{du}{dx}=e^u \cdot 2 = 2e^{2x}
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  3. #3
    Super Member Gamma's Avatar
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    Right on dude. Looks like you got it to me. good luck on the exam.
    \frac{d}{dx}e^{2x}=2e^{2x}
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    Thank you! =D
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  5. #5
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    Quote Originally Posted by Von Lauder View Post
    I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.

    I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.

    However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?

    correct

    If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?

    correct

    Please correct anything I am mistaken on. Thank you.
    u a function of x ...

    \frac{d}{dx} [e^u] = e^u \cdot \frac{du}{dx}


    k a konstant ...

    \int e^{kx} \, dx = \frac{1}{k} e^{kx} + C
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  6. #6
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    Quote Originally Posted by Von Lauder View Post
    I have to take the AP Calculus AB test tomorrow, so anyone who can clarify this for me quickly would be much, much appreciated.

    I understand that when I perform the chain rule with a derivative, such as d/dx (2x+1)^3, I must take the derivative, yielding 3(2x+1)^2, and then I must multiply by the derivative of the inside, d/dx (2x+1) = 2, yielding 2*3(2x+1)^2 -> 6(2x+1)^2.

    However, I am confused on this when I am using something other than a simple binomial. If I must evaluate d/dx (e^2x), would the derivative simply be (e^2x), or would I multiply that by the derivative of the exponent (d/dx 2x = 2), making it 2*(e^2x) -> (2e^2x)?
    Yes, any time you have a "differentiation rule" for f(x) but have some function of x, say u(x), you must use the chain rule:
    \frac{df(u(x))}{dx}= \frac{df(u)}{du}\frac{du}{dx}
    In this particular example, u(x)= 2x so
    \frac{de^u}{dx}= \frac{de^u}{du}\frac{du}{dx}= \frac{e^u}{du}\frac{2x}{dx}= (e^u)(2)= 2e^{2x}

    A slightly more complicated example is:
    \frac{d e^{x^2}}{dx}
    Here u(x)= x^2 so
    \frac{d e^{x^2}}{dx}= \frac{de^u}{du}\frac{du}{dx}= \frac{de^u}{du}\frac{d(x^2)}{dx}= (e^u)(2x)= 2xe^{x^2}

    If the latter is true and I must multiply by the derivative of the exponent, then what do I do when I must take the antiderivative of it? For example, if I am given ∫(e^2x)dx, would the answer be (1/2)(e^2x)+C ?
    That's exactly what "substitution" in integration is. To integrate \int e^{2x}dx[/tex], let u= 2x. then du= 2dx so (1/2)du= dx. Replacing 2x by u and dx by (1/2)du, the integral becomes
    [tex]\int e^{2x}dx= \int e^u ((1/2)du)= (1/2)\int e^u du= (1/2)e^u+ C= (1/2)e^{2x}+ C[//math]
    just as you say.

    HOWEVER, integrating is harder than differentiating and this is one reason. If the integral were
    \int e^{x^2} dx
    we might try the substitution u= x^2. Then du= (1/2)x dx. If we do the same as before, 2du/x= dx and when we replace x^2 by u and dx by 2du/x we get
    \int 2 e^u/x du
    and still have x in the integral. We can't take it out of the integral as we did with the "2" in the first integral because it is a variable and not a constant. That is why
    \int e^{-x^2} dx
    important in the "Normal Probability Distribution", cannot be written in terms of elementary functions.

    Notice that, even though it seems more difficult, the integral
    \int xe^{x^2}dx
    CAN be done easily. Using the substitution u= x^2 now, du= 2x dx, [tex] (1/2)du= x dx and the integral becomes
    (1/2)\int e^u du= (1/2)e^u+ C= (1/2)e^{x^2}+ C

    Please correct anything I am mistaken on. Thank you.
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  7. #7
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    Possible Correction.

    Quote Originally Posted by HallsofIvy View Post
    HOWEVER, integrating is harder than differentiating and this is one reason. If the integral were
    \int e^{x^2} dx
    we might try the substitution u= x^2. Then du= (1/2)x dx. If we do the same as before, 2du/x= dx and when we replace x^2 by u and dx by 2du/x we get
    \int 2 e^u/x du
    and still have x in the integral. We can't take it out of the integral as we did with the "2" in the first integral because it is a variable and not a constant.
    If u=x^2, then I think that du = 2x dx, not du = \frac{1}{2}x dx. I think what you meant was that dx = \frac{1}{2x}du.

    Either way, you are stuck with two variables in the integral. I just thought I'd point that out. (And there is a good chance that you are right, and I am wrong by the way).
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