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Math Help - Differentiate

  1. #1
    Member Awsom Guy's Avatar
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    Differentiate

    Differentiate this function f(x)= (4x)/(sqrt2x).
    I know I use the quotient rule to differentiate this, this is what I have done so far:

    (2x)^1/2 (0) - (4) [1/2(2x)^-1/2] divided by (2x)

    0 - 2(2x)^-1/2 divided by (2x)

    -(2x)^-1/2 divided by x

    -2 divided by x^1/2 * x

    -2 divided by sqrtx^3

    Please help if something here is confusing because i dont know how to use Latex please tell I will try to show it more better.
    Thanks
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  2. #2
    Member Nacho's Avatar
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    fastter than that, is note that: f(x)=\dfrac{4}{\sqrt{2}}\cdot \dfrac{x}{\sqrt{x}}
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  3. #3
    Member Awsom Guy's Avatar
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    but what I have done is not right. Tha answer given is (2)/(sqrt2x)
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  4. #4
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    Quote Originally Posted by Nacho View Post
    fastter than that, is note that: f(x)=\dfrac{4}{\sqrt{2}}\cdot \dfrac{x}{\sqrt{x}}
    Nacho is pointing out that his approach and hint will work better. Instead, split it into a constant ( \frac{4}{\sqrt 2}) multiplied by a term in x ( \frac{x}{\sqrt x}=\sqrt x)

    So then you can just use the power rule. So before differentiating you have changed the function to y=\frac{4\sqrt x}{\sqrt2}

    Can you differentiate that more easily?
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  5. #5
    Member Nacho's Avatar
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    ok, I donīt undestand why you multiplicate by zero in the first line, if the quotient rule says: <br />
\left( {\frac{h}<br />
{g}} \right)^\prime   = \frac{{h'g - hg'}}<br />
{{g^2 }}<br />

    In your example h=4x and g=\sqrt{2x}

    Whit my way is easy <br />
f(x) = \frac{4}<br />
{{\sqrt 2 }} \cdot x^{1/2}  \Rightarrow f'(x) = \frac{4}<br />
{{\sqrt 2 }} \cdot \frac{1}<br />
{2} \cdot x^{ - 1/2}  = \frac{{\sqrt 2 }}<br />
{{\sqrt x }}<br />
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  6. #6
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    Quote Originally Posted by Nacho View Post
    ok, I donīt undestand why you multiplicate by zero in the first line, if the quotient rule says: <br />
\left( {\frac{h}<br />
{g}} \right)^\prime   = \frac{{h'g - hg'}}<br />
{{g^2 }}<br />

    In your example h=4x and g=\sqrt{2x}

    Whit my way is easy <br />
f(x) = \frac{4}<br />
{{\sqrt 2 }} \cdot x^{1/2}  \Rightarrow f'(x) = \frac{4}<br />
{{\sqrt 2 }} \cdot \frac{1}<br />
{2} \cdot x^{ - 1/2}  = \frac{{\sqrt 2 }}<br />
{{\sqrt x }}<br />
    Just pointing out simplification

    \frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}
    Last edited by Keithfert488; February 12th 2010 at 09:16 PM.
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  7. #7
    Member Nacho's Avatar
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    Quote Originally Posted by Keithfert488 View Post
    Just pointing out an error in simplification

    \frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}
    I just continue the simplification

    <br />
\frac{2}<br />
{{\sqrt 2 }} = \frac{{\sqrt 2 \sqrt 2 }}<br />
{{\sqrt 2 }} = \sqrt 2 <br />
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  8. #8
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    Quote Originally Posted by Keithfert488 View Post
    Just pointing out an error in simplification

    \frac{4}{\sqrt{2}}\cdot \frac{1}{2}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2}}\cdot x^{-\frac{1}{2}}=\frac{2}{\sqrt{2x}}
    Your solution and his solution are equal.
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  9. #9
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    Quote Originally Posted by drumist View Post
    Your solution and his solution are equal.
    Wow. I feel stupid. Well...I was doing it so we can see it through to the answer that he gave us.
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  10. #10
    Member Awsom Guy's Avatar
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    lols thanks guys.
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