# Proving $\frac{d}{dx}x^2=2x$ by definition

I did the following proof earlier and just wanted conformation as to whether it works. The question was to show $$\frac{d}{dx}x^2=2x$$ by the difference-quotient definition of a derivative, and then prove that limit with the $\epsilon$, $\delta$ definition for limits.

I said that $$\frac{d}{dx}x^2=\lim_{h\to0}\frac{(x+h)^2-x^2}{h}=\lim_{h\to0}\frac{h^2+2xh}{h}$$ Then we assert that this is equal to $2x$ which we show is true using the $\epsilon$, $\delta$ definition. We must show that for all $\epsilon > 0$, there exists $\delta>0$ such that for all $|h-0|<\delta$ we have $$\left|2x-\frac{h^2+2xh}{h}\right|<\epsilon$$ Assuming $h\neq 0$ this simplifies to $$\left|2x-h-2x\right|=|-h|=|h|<\epsilon$$ So we're left with that for all $|h|<\delta$ we must have $|h|<\epsilon$, which is true for all $\epsilon$ if $\delta=\epsilon$. Hence we can conclude the limit does equal $2x$ and $$\frac{d}{dx}x^2=2x$$
I know its kind of a dull question, so if you have any cool insights or tips to share that'd be appreciated!

• Maybe try to prove $$\frac{d}{dx} x^n =nx^{n-1}$$ for $n \in \mathbb{N}$? :) Commented May 13, 2016 at 1:18

Yes it is correct. A tip is to immediately write: $$\lim_{h \rightarrow 0} \frac{h^2+2xh}{h}=\lim_{h \rightarrow 0} 2x+h.$$ You can now see directly, to get $2x+h$ within $\epsilon>0$ of $2x$, we must choose $|h|<\epsilon$.
Note that if $f$ is a continuous function, and $f(0)$ exists, then $\lim_{h\to 0} f(h) = f(0)$ (this is just the definition of continuity).
This lets you immediately plug in $h=0$ in after dividing by $h$, avoiding the epsilon-delta portion of your argument.