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anonymous

  • one year ago

Using the difference quotient, find f'(x) from 2sqrtx

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  1. anonymous
    • one year ago
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    I know the answer is \[\frac{ 1 }{ \sqrt{x} }\]

  2. anonymous
    • one year ago
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    But I get only as far as \[\frac{ 2x + 2h + 2x}{ 2\sqrt{x} + 2\sqrt{x}}\]

  3. Jhannybean
    • one year ago
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    ok so difference quotient: \(\dfrac{f(x+h)-f(x)}{h}\) I would find what \(f(x+h)\) is and state what \(f(x)\) is separately, then just plug it into the formula. \[f(x) = 2\sqrt{x}\]\[f(x+h) = 2\sqrt{x+h}\]\[\begin{align} f'(x)= \dfrac{2\sqrt{x+h}-2\sqrt{x}}{h} \\ &=\frac{2\sqrt{x+h} -2\sqrt{x}}{h} \cdot \frac{2\sqrt{x+h}+2\sqrt{x}}{2\sqrt{x+h}+2\sqrt{x}} \\ &=\frac{4(x+h)-4x}{h(2\sqrt{x+h}+2\sqrt{x})} \\&=\frac{4h}{h(2\sqrt{x+h}+2\sqrt{x})} \\ &=\color{red}{\frac{4}{2\sqrt{x+h}+2\sqrt{x}}} \end{align} \]

  4. anonymous
    • one year ago
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    THANK YOU

  5. Jhannybean
    • one year ago
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    did you follow?

  6. anonymous
    • one year ago
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    Yes I messed up I saw my mistake you da bestest!

  7. Jhannybean
    • one year ago
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    Remember that: \((a-b)(a+b) = a^2-b^2\)

  8. anonymous
    • one year ago
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    thanks i follow....:)

  9. Jhannybean
    • one year ago
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    Awesome

  10. anonymous
    • one year ago
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    i love u

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