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Math Help - Differentiating E

  1. #1
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    Differentiating E

    I'm really not understanding how to differentiate things when E to the power of something is present.

    For example:

    Differentiate:

    X^2 times e^-x

    Apply Product Rule:

    sol'n: 2xe^-x + x^2 ??

    The e^-x is confusing me.
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  2. #2
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    Quote Originally Posted by Xavier20 View Post
    I'm really not understanding how to differentiate things when E to the power of something is present.

    For example:

    Differentiate:

    X^2 times e^-x

    Apply Product Rule:

    sol'n: 2xe^-x + x^2 ??

    The e^-x is confusing me.
    let u be a function

    d/du e^u = u' e^u

    in other words, to differentiate e raised to any power, you multiply e raised to the original power by the derivative of the power.

    example:

    d/dx e^(x^2) = 2xe^(x^2) ..........2x is the derivative of x^2

    now try the question again
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  3. #3
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    X^2 times e^-x

    2xe^-x + x^2 times -xe^-x

    ?
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    Quote Originally Posted by Xavier20 View Post
    X^2 times e^-x

    2xe^-x + x^2 times -xe^-x

    ?
    no, the derivative of -x is -1 not -x

    try one more time
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  5. #5
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    Ohh. Right.

    2xe^-x + x^2 times -e^-x
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    Quote Originally Posted by Xavier20 View Post
    Ohh. Right.

    2xe^-x + x^2 times -e^-x
    yeah, but you wouldn't write "times". write it out like a mathematician
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  7. #7
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    2xe^-x + (x^2)(-e^-x)
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    Quote Originally Posted by Xavier20 View Post
    2xe^-x + (x^2)(-e^-x)
    well that's better, but this is even better

    2xe^-x - (x^2)e^-x

    or to factorize:

    (xe^-x)(2 - x)
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  9. #9
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    try these (if you have time, if you're doing homework otherwise, don't bother):

    Find the derivatives of the following:

    y = e^(ax^3) .........where a is a constant

    y = e^(tsin(2t))

    y = sqrt(1 + 2e^(3x))

    y = e^(e^x)

    y = cos(e^(pi*x))
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  10. #10
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    Find the derivative of the following:

    y = e^(ax^3) .........where a is a constant

    y' = (3ax^2)e^(ax^3)

    y = e^(tsin(2t))

    y' = [sin(2t)tcos(2t)]e^(tsin(2t))

    y = sqrt(1 + 2e^(3x))

    y = (1+2e^(3x))^1/2
    y' = [1/2(1+2e^(3x))^-1/2]6e^(3x)

    y = e^(e^x)

    y' = (e^x)e^(e^x)

    y = cos(e^(pi*x))

    y' = [-sin(e^(pi*x))]pi*e^(pi*x)
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  11. #11
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by Xavier20 View Post

    y = e^(tsin(2t))

    y' = [sin(2t)tcos(2t)]e^(tsin(2t))
    Good! all were correct except the one above, try to see where you made the mistake
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  12. #12
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    y = e^(tsin(2t))

    y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

    I forgot the chain rule part of cos(2t).
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  13. #13
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    Quote Originally Posted by Xavier20 View Post
    y = e^(tsin(2t))

    y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

    I forgot the chain rule part of cos(2t).
    you need both the product and chain rule to find the derivative of tsin(2t)

    you have 2 functions in a product, one is t and one is sin(2t) so you need the product rule. but sin(2t) is a composite function, so you need the chain rule for that

    Product rule:

    d/dx f(x)*g(x) = f ' (x)*g(x) + g'(x)*f(x)

    Chain rule:

    d/dx f(g(x)) = f ' (g(x))*g'(x)

    here goes:

    d/dx tsin(2t) = sin(2t) [i took the derivative of t here and multiplied by sin(2t)] + t(2cos(2t)) [i left the t and took the derivative of sin(2t) using the chain rule]

    so d/dx tsin(2t) = sin(2t) + 2tcos(2t)

    now you can complete the problem
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  14. #14
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    Looks like you got trouble with "e" number

    Always remember that



    Could be you are messin' it with chain rule.
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  15. #15
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    y' = [sin(2t)tcos(2t)*2]e^(tsin(2t))

    Yeah. I forgot to add the + sign between the sin(2t) and tcos(2t) and didn't bring the *2 out infront of the tcos(2t) in my chain rule.

    Thanks for sorting that out though.

    It should be:

    y' = [sin(2t) + 2tcos(2t)]e^(tsin(2t))
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