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Thread: simple diff quesstion

  1. #1
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    simple diff quesstion

    want to check a question and solution.


    differentiate with respect to x

    $\displaystyle y= (x^3+4)/x$

    = $\displaystyle x^2+4x^-{^1}$
    ( this is the line im confused about why does the x^3 become x^2 and not 3x) or is it wrong.

    $\displaystyle dy/dx = 2x-4x^{-2}$

    many thanks
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    Quote Originally Posted by decoy808 View Post
    want to check a question and solution.


    differentiate with respect to x

    $\displaystyle y= (x^3+4)/x$

    = $\displaystyle x^2+4x^-{^1}$
    ( this is the line im confused about why does the x^3 become x^2 and not 3x) or is it wrong.

    $\displaystyle dy/dx = 2x-4x^{-2}$

    many thanks
    Note that $\displaystyle \frac{x^3 + 4}{x} = \frac{x^3}{x} + \frac{4}{x} = x^2 + 4x^{-1}$.


    So when you take the derivative, it will be $\displaystyle 2x - 4x^{-2} = 2x - \frac{4}{x^2}$.
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    Quote Originally Posted by decoy808 View Post
    want to check a question and solution.


    differentiate with respect to x

    $\displaystyle y= (x^3+4)/x$

    = $\displaystyle x^2+4x^-{^1}$
    ( this is the line im confused about why does the x^3 become x^2 and not 3x) or is it wrong
    Yes, that is correct. It is $\displaystyle \frac{x^4+ 4}{x}= \frac{x^4}{x}+ \frac{4}{x}$. The "$\displaystyle x^3$" becomes "$\displaystyle x^2$" because the x in the denominator cancels one x in the numerator.

    $\displaystyle dy/dx = 2x-4x^{-2}$

    many thanks
    Yes, that derivative is correct.

    Two minutes late again! I need to type faster!
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    how do i do y= (x+1)/x
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    Quote Originally Posted by decoy808 View Post
    how do i do y= (x+1)/x
    $\displaystyle \frac{x + 1}{x} = \frac{x}{x} + \frac{1}{x} = 1 + x^{-1}$.


    So the derivative is $\displaystyle -1x^{-2} = -\frac{1}{x^2}$.
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