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

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frecklesBest ResponseYou've already chosen the best response.2
hey you left something off man on x_0
 one year ago

frecklesBest ResponseYou've already chosen the best response.2
where you have f(x_0)
 one year ago

satellite73Best ResponseYou've already chosen the best response.2
\[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2x_0^2) \\ & \end{align}\]
 one year ago

frecklesBest ResponseYou've already chosen the best response.2
gosh golly where you see x_0 it should be x_0^2
 one year ago

frecklesBest ResponseYou've already chosen the best response.2
where you have f(x_0) you suppose to have x_0^2
 one year ago

frecklesBest ResponseYou've already chosen the best response.2
sat left is off in his middle two lines
 one year ago

zeppBest ResponseYou've already chosen the best response.0
So \[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2x_0^2}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2x_0^2) \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(2x_0\Delta x+\Delta x^2) \\ &=\lim_{\Delta x \rightarrow 0}(2x_0+\Delta x) \end{align}\] ?
 one year ago

frecklesBest ResponseYou've already chosen the best response.2
you missed it in the third line but yeah you got it lol
 one year ago

zeppBest ResponseYou've already chosen the best response.0
I replace Delta x by 0, then I get \(2x_0\), that 0 is unecessary so \[\frac{d}{dx}x^2=2x\]
 one year ago

frecklesBest ResponseYou've already chosen the best response.2
Yes! Brilliant! Good job! :)
 one year ago
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