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Thread: Need help with finding derivatives

  1. #1
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    Need help with finding derivatives

    y= sqrt[x] / 2sqrt[x]+3


    y=-2sqrt[x^2+4x]
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  2. #2
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    Can you use any method to solve or do you need to use the fundamental theorem?
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  3. #3
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    Is the first one $\displaystyle \displaystyle y = \frac{\sqrt{x}}{2\sqrt{x}} + 3$ or $\displaystyle \displaystyle y = \frac{\sqrt{x}}{2\sqrt{x} + 3}$?
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  4. #4
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    Quote Originally Posted by Prove It View Post
    Is the first one $\displaystyle \displaystyle y = \frac{\sqrt{x}}{2\sqrt{x}} + 3$ or $\displaystyle \displaystyle y = \frac{\sqrt{x}}{2\sqrt{x} + 3}$?
    the second one can you answer please ?
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  5. #5
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    have you tried the quotient rule for derivatives?
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    Well first of all, since you're dividing by ALL of $\displaystyle \displaystyle 2\sqrt{x} + 3$, you needed to write it as sqrt[x]/(2sqrt[x] + 3)...

    Now you should apply the Quotient Rule, namely if $\displaystyle \displaystyle u = u(x)$ and $\displaystyle \displaystyle v = v(x) \neq 0$, then

    $\displaystyle \displaystyle \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v\,\frac{du}{dx} - u\,\frac{dv}{dx}}{v^2}$.
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  7. #7
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    yeah im just having trouble with the square roots though . like do i get rid of them ?
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  8. #8
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    Remember that $\displaystyle \displaystyle \sqrt{x} = x^{\frac{1}{2}}$. Then you can apply the power rule for derivatives...

    BTW, since this is clearly a Calculus question, why have you posted in Pre-Calculus?
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  9. #9
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    thanks for the help
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  10. #10
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    taking the derivative of functions with $\displaystyle \sqrt{x}$ in them are much easier if you remember that its derivative is $\displaystyle \dfrac{1}{2\sqrt{x}}$.

    quotient rule ...

    $\displaystyle y' = \dfrac{(2\sqrt{x}+3) \cdot \dfrac{1}{2\sqrt{x}} - \sqrt{x} \cdot \dfrac{1}{\sqrt{x}}}{(2\sqrt{x}+3)^2}$

    multiply stuff in the numerator and combine like terms ...

    $\displaystyle y' = \dfrac{\dfrac{3}{2\sqrt{x}}}{(2\sqrt{x}+3)^2}$

    clean it up ...

    $\displaystyle y' = \dfrac{3}{2\sqrt{x}(2\sqrt{x}+3)^2}$
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  11. #11
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    Quote Originally Posted by skeeter View Post
    taking the derivative of functions with $\displaystyle \sqrt{x}$ in them are much easier if you remember that its derivative is $\displaystyle \dfrac{1}{2\sqrt{x}}$.

    quotient rule ...

    $\displaystyle y' = \dfrac{(2\sqrt{x}+3) \cdot \dfrac{1}{2\sqrt{x}} - \sqrt{x} \cdot \dfrac{1}{\sqrt{x}}}{(2\sqrt{x}+3)^2}$

    multiply stuff in the numerator and combine like terms ...

    $\displaystyle y' = \dfrac{\dfrac{3}{2\sqrt{x}}}{(2\sqrt{x}+3)^2}$

    clean it up ...

    $\displaystyle y' = \dfrac{3}{2\sqrt{x}(2\sqrt{x}+3)^2}$
    Not necessarily... it's just another formula to remember ><
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