Question

Derivative of \(x^2\)?

I used the difference quotient formula, but I'm stuuuuck :(

\[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\
&=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0}{\Delta x} \\
&=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\
&=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0) \\
&=\lim_{\Delta x \rightarrow 0}(\frac{x_0^2}{\Delta x}+2x_0+\frac{1}{\Delta x}-\frac{x_0}{\Delta x}) \end{align}\]

Answer 1

hey you left something off man on x_0

Answer 2

\[f(x_0)=x_0^2\]

Answer 3

Where?

Answer 4

where you have f(x_0)

Answer 5

\[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\
&=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0}{\Delta x} \\
&=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\
&=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0^2) \\
& \end{align}\]

Answer 6

gosh golly where you see x_0 it should be x_0^2

Answer 7

Ohh.

Answer 8

where you have f(x_0) you suppose to have x_0^2

Answer 9

sat left is off in his middle two lines

Answer 10

So

\[\large \begin{align}\frac{d}{dx}x^2&=\lim_{\Delta x \rightarrow 0}\frac{f(x_0+\Delta x)-f(x_0)}{\Delta x}\\
&=\lim_{\Delta x \rightarrow 0}\frac{(x_0+\Delta x)^2-x_0^2}{\Delta x} \\
&=\lim_{\Delta x \rightarrow 0}\frac{x_0^2+2x_0\Delta x+\Delta x^2-x_0}{\Delta x} \\
&=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(x_0^2+2x_0\Delta x+\Delta x^2-x_0^2) \\ &=\lim_{\Delta x \rightarrow 0}\frac{1}{\Delta x}(2x_0\Delta x+\Delta x^2) \\
&=\lim_{\Delta x \rightarrow 0}(2x_0+\Delta x) \end{align}\]
?

Answer 11

you missed it in the third line but yeah you got it lol

Answer 12

I replace Delta x by 0, then I get \(2x_0\), that 0 is unecessary so \[\frac{d}{dx}x^2=2x\]

Answer 13

Thanks! :)

Answer 14

Yes! Brilliant! Good job! :)