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Math Help - Slope of tangent line

  1. #1
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    Slope of tangent line

    I have to find the derivative of this problem but I can't figure the algebra out. I can't use the power rule either. I have to use the (f(x+h)-f(x))/h rule.

    1/x^(1/2)
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  2. #2
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    Quote Originally Posted by TenaciousE View Post
    I have to find the derivative of this problem but I can't figure the algebra out. I can't use the power rule either. I have to use the (f(x+h)-f(x))/h rule.

    1/x^(1/2)
    \displaystyle\lim_{h\to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}
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  3. #3
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    @dwsmith I have that much figured out. I'm having trouble simplifying it to find the equation for the slope of the tangent line
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    \displaystyle \frac{1}{h}\left(\sqrt{x+h}-\sqrt{x}\right)

    \displaystyle \frac{1}{h}\left(\sqrt{x+h}-\sqrt{x}\times \frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}}\ri  ght)

    Have a go from here...
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  5. #5
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    Quote Originally Posted by TenaciousE View Post
    @dwsmith I have that much figured out. I'm having trouble simplifying it to find the equation for the slope of the tangent line
    \displaystyle\lim_{h\to 0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{x\to 0}\frac{\frac{\sqrt{x}-\sqrt{x+h}}{\sqrt{x}\sqrt{x+h}}}{h}=\lim_{x\to 0}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x}\sqrt{x+h}}

    Multiple by the conjugate.
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  6. #6
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    Got it figured out now. Thanks!
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