zepp
  • zepp
Derivative of \(x^2\)? I used the difference quotient formula, but I'm stuuuuck :( \[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0) \\ &=\lim_{\Delta x \rightarrow 0}(\frac{x_0^2}{\Delta x}+2x_0+\frac{1}{\Delta x}-\frac{x_0}{\Delta x}) \end{align}\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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freckles
  • freckles
hey you left something off man on x_0
freckles
  • freckles
\[f(x_0)=x_0^2\]
zepp
  • zepp
Where?

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freckles
  • freckles
where you have f(x_0)
anonymous
  • anonymous
\[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0^2) \\ & \end{align}\]
freckles
  • freckles
gosh golly where you see x_0 it should be x_0^2
zepp
  • zepp
Ohh.
freckles
  • freckles
where you have f(x_0) you suppose to have x_0^2
freckles
  • freckles
sat left is off in his middle two lines
zepp
  • zepp
So \[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0^2}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0^2) \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(2x_0\Delta x+\Delta x^2) \\ &=\lim_{\Delta x \rightarrow 0}(2x_0+\Delta x) \end{align}\] ?
freckles
  • freckles
you missed it in the third line but yeah you got it lol
zepp
  • zepp
I replace Delta x by 0, then I get \(2x_0\), that 0 is unecessary so \[\frac{d}{dx}x^2=2x\]
zepp
  • zepp
Thanks! :)
freckles
  • freckles
Yes! Brilliant! Good job! :)

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