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

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

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

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

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

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

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

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

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

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