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Math Help - Quotient / Product Rule help

  1. #1
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    Quotient / Product Rule help

    I need help differentiating this:

    y = [ 7+xf(x) ] / sqrt(x)

    Thanks!
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    Quote Originally Posted by moblicious View Post
    I need help differentiating this:

    y = [ 7+xf(x) ] / sqrt(x)

    Thanks!
    How did you begin? where are you stuck? you know what the quotient and product rules say, right?

    \dsiplaystyle y' = \frac {\sqrt x \cdot \frac d{dx}(7 + xf(x)) - (7 + xf(x)) \cdot \frac d{dx} \sqrt x}{(\sqrt x)^2}

    That's plugging it into the quotient rule. now what?
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    Quote Originally Posted by Jhevon View Post
    How did you begin? where are you stuck? you know what the quotient and product rules say, right?

    \dsiplaystyle y' = \frac {\sqrt x \cdot \frac d{dx}(7 + xf(x)) - (7 + xf(x)) \cdot \frac d{dx} \sqrt x}{(\sqrt x)^2}

    That's plugging it into the quotient rule. now what?
    I thought I had to use the product rule to find the numerator first, then use the quotient rule to find the remaining numbers.

    So that's plugged in the quotient rule, would it equal to:

    f'(x)-sqrt(x)-(7+x(fx))-1/2x^(-1/2) / x?
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by moblicious View Post
    I thought I had to use the product rule to find the numerator first, then use the quotient rule to find the remaining numbers.

    So that's plugged in the quotient rule, would it equal to:

    f'(x)-sqrt(x)-(7+x(fx))-1/2x^(-1/2) / x?
    yes, you need to use the product rule to find the derivative of xf(x). but you start applying the quotient rule first, and then realize in the process that you need to apply the product rule.

    and no, that's not what the derivative would be.
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    f(x)+xf'(x)sqrt(x)-(7+f(x)+xf'(x))-1/2x^(-1/2) / x?
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    Quote Originally Posted by moblicious View Post
    f(x)+xf'(x)sqrt(x)-(7+f(x)+xf'(x))-1/2x^(-1/2) / x?
    first, i don't get what you're writing. learn LaTeX or use parentheses properly. the derivative is:

    \displaystyle y' = \frac {\sqrt x [f(x) + xf'(x)] - [7 + xf(x)] \cdot \frac 12x^{-1/2}}x

    Now simplify
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  7. #7
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    I used every parentheses correctly, don't know what you were looking at.

    Is it (xf(x)+2x^2f'(x)-7) / 2x^(3/2) ?
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  8. #8
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by moblicious View Post
    I used every parentheses correctly
    I can say with certainty that you didn't. for one, the numerator and denominator should be encapsulated in parentheses when typing like that. if you look at what you typed, this was not the case; the "f(x)+xf'(x)" should have been in parentheses also... anyway

    Is it (xf(x)+2x^2f'(x)-7) / 2x^(3/2) ?
    yes, this is correct
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  9. #9
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    My apologies then. Thanks for the help.
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  10. #10
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by moblicious View Post
    My apologies then. Thanks for the help.
    no worries. good luck.
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