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Math Help - definition of derivative

  1. #1
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    definition of derivative

    Using the definition of derivative how do i solve this equation: f(x)=5/square root 3x. Please help....? Ive used the fast short cuts to find this derivative but I cant seem to get how to do it by the definition way and i have a test over definition tomorrow. Please help me. im desperate. My answer that i got not using defintion of -5/2x sqroot3x. My tutor couldnt even figure this one out.
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  2. #2
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    Also my tutor and I were trying to use the conjugate rule and we thought we were getting close but no luck.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by frankmiller View Post
    Using the definition of derivative how do i solve this equation: f(x)=5/square root 3x. Please help....? Ive used the fast short cuts to find this derivative but I cant seem to get how to do it by the definition way and i have a test over definition tomorrow. Please help me. im desperate. My answer that i got not using defintion of -5/2x sqroot3x. My tutor couldnt even figure this one out.
    I'm just making sure I interpreted this correctly. Is the function f\!\left(x\right)=\frac{5}{\sqrt{3x}}?
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  4. #4
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    yes that is correct.
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  5. #5
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    Can you simplify \frac{\frac{5}{{\sqrt {x + h} }} - \frac{5}{{\sqrt x }}}{h}?

    Then what happens if h \to 0 ~?
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  6. #6
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    so you can pull the 3 out of the top? I wasnt sure.
    From there i would do:
    [tex](5)(sqrt x)/sqrt(x+h)(sqrt x)-(5)sqrt(x+h)/sqrt(x+h)(sqrt x) /h

    But even here im confused. I try to take an h out to remove the bottom denominator but then how will h approach 0?
    sorry if im being confusing. Im just really frustrated because this is really the only concept im not getting and my test is tomorrow..
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  7. #7
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by frankmiller View Post
    so you can pull the 3 out of the top? I wasnt sure.
    From there i would do:
    [tex](5)(sqrt x)/sqrt(x+h)(sqrt x)-(5)sqrt(x+h)/sqrt(x+h)(sqrt x) /h

    But even here im confused. I try to take an h out to remove the bottom denominator but then how will h approach 0?
    sorry if im being confusing. Im just really frustrated because this is really the only concept im not getting and my test is tomorrow..
    Following Plato's suggestion, that difference quotient becomes 5\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^2+hx}}

    Now apply the conjugate trick.
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  8. #8
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    Im really not good at the conjugate rule. could you give me a little help.
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  9. #9
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    actually hold on a sec. I might have it.
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  10. #10
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    my work so far:
    5 sqrt(x)-sqrt(x+h) (sqrt(x)-sqrt(x+h)/h sqrt(x^2+hx) qrt(x)-sqrt(x+h)

    but im stuck here:

    5(x-x+h)/h sqrt(x^2+hx) qrt(x)-sqrt(x+h)
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  11. #11
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    wait so then
    5h/h sqrt(x^2+hx)sqrt(x)-sqrt(x+h)

    ??? anyone
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  12. #12
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by frankmiller View Post
    Im really not good at the conjugate rule. could you give me a little help.
    Quote Originally Posted by Chris L T521 View Post
    Following Plato's suggestion, that difference quotient becomes 5\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^2+hx}}

    Now apply the conjugate trick.
    From here, we have

    \begin{aligned}5\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^2+hx}}\cdot\frac{\sqrt{x}+\sq  rt{x+h}}{\sqrt{x}+\sqrt{x+h}} & = 5\frac{x-(x+h)}{h\sqrt{x^2+hx}\left(\sqrt{x}+\sqrt{x+h}\rig  ht)}\\ & = -5\frac{1}{\sqrt{x^2+hx}\left(\sqrt{x}+\sqrt{x+h}\r  ight)}\end{aligned}

    Now, when we evaluate \lim_{h\to0}-5\frac{1}{\sqrt{x^2+hx}\left(\sqrt{x}+\sqrt{x+h}\r  ight)}, we have -5\frac{1}{2\sqrt{x^2}\sqrt{x}}=-\frac{5}{2\sqrt{x^3}}.

    Keep in mind, though that we ignored the \sqrt{3} term in the original problem.

    So we can conclude then that if f\!\left(x\right)=\frac{5}{\sqrt{3x}}, then f^{\prime}\!\left(x\right)=-\frac{5}{2\sqrt{3}\sqrt{x^3}}=-\frac{5}{2\sqrt{3}}x^{-\frac{3}{2}}, which is what we would expect when differentiating using the shortcuts.

    Does this make sense?
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