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

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freckles Group TitleBest ResponseYou've already chosen the best response.2
hey you left something off man on x_0
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.2
\[f(x_0)=x_0^2\]
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.2
where you have f(x_0)
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.2
\[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)f(x_0)}{\Delta x}\\ &=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2x_0}{\Delta x} \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2x_0^2) \\ & \end{align}\]
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.2
gosh golly where you see x_0 it should be x_0^2
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.2
where you have f(x_0) you suppose to have x_0^2
 2 years ago

freckles Group TitleBest ResponseYou've already chosen the best response.2
sat left is off in his middle two lines
 2 years ago

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

freckles Group TitleBest ResponseYou've already chosen the best response.2
you missed it in the third line but yeah you got it lol
 2 years ago

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

freckles Group TitleBest ResponseYou've already chosen the best response.2
Yes! Brilliant! Good job! :)
 2 years ago
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