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Math Help - Urgent differential help!!

  1. #1
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    Urgent differential help!!

    Hey guys

    can you help with this please?

    9) Differentiate xe^{\sqrt{x}}


    Please help,

    thankyou guys!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anthmoo View Post
    Hey guys

    can you help with this please?

    9) Differentiate xe^{\sqrt{x}}


    Please help,

    thankyou guys!
    we need the product rule here:

    recall, by the product rule: \frac d{dx}f(x)g(x) = f'(x)g(x) + f(x)g'(x)

    also recall that \frac d{dx}e^u = u'e^u, where u is a function of x


    try it and see what you get
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  3. #3
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    I did...but I keep getting 5e^2

    I think its differentiating e^{\sqrt{x}} that's the problem. Can you help with that?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anthmoo View Post
    I did...but I keep getting 5e^2

    I think its differentiating e^{\sqrt{x}} that's the problem. Can you help with that?
    did you see my last post? when finding the derivative of e to some power, you find the derivative of the power and multiply it by e to the old power

    example: \frac d{dx}e^{x^2} = \frac d{dx}x^2 \cdot e^{x^2} = 2xe^{x^2}

    now try \frac d{dx}e^{\sqrt{x}}
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  5. #5
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    Thankyou so much!

    I can do it now =] all that was needed was that rule you gave for differentiating e to the power of a function. Thank you!

    Anthmoo
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by anthmoo View Post
    Thankyou so much!

    I can do it now =] all that was needed was that rule you gave for differentiating e to the power of a function. Thank you!

    Anthmoo
    so, what's your answer?
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  7. #7
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    I got 2e^2 ...that's right isn't it?

    Thanks =]
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by anthmoo View Post
    I got 2e^2 ...that's right isn't it?

    Thanks =]
    No!! The only way you can get a constant after taking a derivative is if the original function is a constant times x. Does e^{\sqrt{x}} look like cx?

    Recall what Jhevon said:

    Quote Originally Posted by Jhevon View Post
    also recall that \frac d{dx}e^u = u'e^u, where u is a function of x
    Here we have u = \sqrt{x}. So what is u^{\prime}?

    -Dan
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  9. #9
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    Quote Originally Posted by topsquark View Post
    No!! The only way you can get a constant after taking a derivative is if the original function is a constant times x. Does e^{\sqrt{x}} look like cx?

    Recall what Jhevon said:



    Here we have u = \sqrt{x}. So what is u^{\prime}?

    -Dan

    Wait wait!! I'm sorry, the original question asked "if f'(4)"...if it was f'(4)...would that be right??

    Thanks
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  10. #10
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by anthmoo View Post
    Wait wait!! I'm sorry, the original question asked "if f'(4)"...if it was f'(4)...would that be right??

    Thanks
    You tell me.
    u^{\prime} = \frac{1}{2\sqrt{x}}

    -Dan
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