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Math Help - Finding the derivative using the definition (have answer)

  1. #1
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    Finding the derivative using the definition (have answer)

    Find the derivative using the definition (NOT shortcuts, we must use definition).

    y=1/sqrt (x)

    My friend and I got it down to x^(-1) - x^(-3/2)
    but the answer is: -1/2x^(3/2)

    How did they get that? (not using shortcuts)
    Thanks!
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  2. #2
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    \frac{{\frac{1}<br />
{{\sqrt {x + h} }} - \frac{1}<br />
{{\sqrt x }}}}<br />
{h} = \frac{{\sqrt x  - \sqrt {x + h} }}<br />
{{h\sqrt x \left( {\sqrt {x + h} } \right)}} = \frac{{ - h}}<br />
{{h\sqrt x \left( {\sqrt {x + h} } \right)\left( {\sqrt x  + \sqrt {x + h} } \right)}}
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  3. #3
    Super Member Random Variable's Avatar
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     \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x}

     = \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} -  \frac{1}{\sqrt{x}}}{\Delta x} \cdot \frac{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}

     = \lim_{\Delta x \to 0} \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x \Big(\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}\Big)}

     = \lim_{\Delta x \to 0} \frac{\frac{-1}{x(x+ \Delta x)}}{\frac{1}{\sqrt{x+\Delta x}} +  \frac{1}{\sqrt{x}}}

     = \frac{\frac{-1}{x^{2}}}{\frac{2}{\sqrt{x}}} = \frac{-1}{2x^{3/2}}
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  4. #4
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    Quote Originally Posted by Random Variable View Post
     \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} - \frac{1}{\sqrt{x}}}{\Delta x}

     = \lim_{\Delta x \to 0} \frac{\frac{1}{\sqrt{x+\Delta x}} -  \frac{1}{\sqrt{x}}}{\Delta x} \cdot \frac{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}}

     = \lim_{\Delta x \to 0} \frac{\frac{1}{x+\Delta x}-\frac{1}{x}}{\Delta x \Big(\frac{1}{\sqrt{x+\Delta x}} + \frac{1}{\sqrt{x}}\Big)}

     = \lim_{\Delta x \to 0} \frac{\frac{-1}{x(x+ \Delta x)}}{\frac{1}{\sqrt{x+\Delta x}} +  \frac{1}{\sqrt{x}}}

     = \frac{\frac{-1}{x^{2}}}{\frac{2}{\sqrt{x}}} = \frac{-1}{2x^{3/2}}

    THANK YOU!! You're a genius lol. I'm so happy I know how to do this now. :]
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