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zepp Group Title

Derivative of \(x^2\)? I used the difference quotient formula, but I'm stuuuuck :( \[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0) \\ &=\lim_{\Delta x \rightarrow 0}(\frac{x_0^2}{\Delta x}+2x_0+\frac{1}{\Delta x}-\frac{x_0}{\Delta x}) \end{align}\]

  • 2 years ago
  • 2 years ago

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  1. freckles Group Title
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    hey you left something off man on x_0

    • 2 years ago
  2. freckles Group Title
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    \[f(x_0)=x_0^2\]

    • 2 years ago
  3. zepp Group Title
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    Where?

    • 2 years ago
  4. freckles Group Title
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    where you have f(x_0)

    • 2 years ago
  5. satellite73 Group Title
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    \[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0^2) \\ & \end{align}\]

    • 2 years ago
  6. freckles Group Title
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    gosh golly where you see x_0 it should be x_0^2

    • 2 years ago
  7. zepp Group Title
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    Ohh.

    • 2 years ago
  8. freckles Group Title
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    where you have f(x_0) you suppose to have x_0^2

    • 2 years ago
  9. freckles Group Title
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    sat left is off in his middle two lines

    • 2 years ago
  10. zepp Group Title
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    So \[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0^2}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0^2) \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(2x_0\Delta x+\Delta x^2) \\ &=\lim_{\Delta x \rightarrow 0}(2x_0+\Delta x) \end{align}\] ?

    • 2 years ago
  11. freckles Group Title
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    you missed it in the third line but yeah you got it lol

    • 2 years ago
  12. zepp Group Title
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    I replace Delta x by 0, then I get \(2x_0\), that 0 is unecessary so \[\frac{d}{dx}x^2=2x\]

    • 2 years ago
  13. zepp Group Title
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    Thanks! :)

    • 2 years ago
  14. freckles Group Title
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    Yes! Brilliant! Good job! :)

    • 2 years ago
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