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I need your help, my question is how to proof that the derivative of $x^2$ is $2x$?

Please I need a clear explanation.

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6 Answers

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What have you tried so far? If you wish to do it from first principles, then $$\frac{d}{dx} x^2 = \lim_{h \to 0} \frac{(x+h)^2-x^2}{h}$$ Try expanding and simplifying this limit

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1st way: Use the product rule and use that $x'=1$. Then $$(x^2)'=(x\cdot x)'=x'\cdot x + x\cdot x'=2\cdot x.$$

2nd way: Use the definition and particularly the difference quotient: $$ f'(x) :=\lim_{h\to 0} \frac{(x+h)^2-x^2}{h}=\lim_{h\to 0} \frac{2hx+h^2}{h}=\lim_{h\to 0} 2x+h = 2x$$

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The derivative of a function gives the rate of change of that function at a certain point. If you do not know that $$\frac{d}{dx}f(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ is the definition of the derivative then asking for a proof of $f'(x)=2x$ given $f(x)=x^2$ is a pointless question. But if you do then,

\begin{align} f'(x)&=\lim_{h\to 0 }\frac{(x+h)^2-x^2}{h} \\ &=\lim_{h\to 0}\frac{(x^2+2xh+h^2)-x^2}{h} \\ &=\lim_{h\to 0}\frac{2xh+h^2}{h} \\ &=\lim_{h\to 0}2x+h \\ &= 2x \end{align}

Or a much quicker way, we use the general formula $$\frac{d}{dx}x^n=nx^{n-1}.$$ In this case let $n=2$ and you wind up with the same answer.

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Take a straight line that passes by the point $(x,x^2)$ of the curve, and a second point close to it, $(x+h,(x+h)^2)$.

The slope of this line is given by the ratio of the $y$ variation over the $x$ variation. $$m=\frac{\Delta y}{\Delta x}=\frac{(x+h)^2-x^2}{(x+h)-x}=\frac{2xh+h^2}h=2x+h.$$

If you make $h$ smaller and smaller, the line converges to the tangent, with slope $2x$.

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The Greeks knew 2 things about parabolas:

  • The focus of a parabola is located at $(0, \frac 14)$
  • The triangle formed from the focus point $F$, to the parabola at $I$, back to the y-axis along the tangent line to $P$, and from $P$ back to $F$ forms an isosceles triangle:

Parabola focus

We can combine these two observations to find the derivative without resorting to limits.

We know:

  • $F = (0,\frac 14)$
  • $I = (x, x^2)$
  • $|IF| = |PF| = L$

So $L = \sqrt{ \left(x^2 - \frac 14\right)^2 + x^2 } = \sqrt{ x^4 + \frac 12x^2 + \frac 1{16} } = x^2 + \frac 14$

From this, we find $P$, as $P = (0, \frac 14 - L) = (0, -x^2)$

Now we can find the slope of the parabola at $I$ by looking the slope of $IP$, which is

$$\begin{align}\text{Slope } IP &= \frac{I_y - P_y}{I_x - P_x}\\ &= \frac{x^2 - -x^2}{x - 0}\\ &= 2x \end{align}$$

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f(x + E) = f(x) + Ef'(x) & E^2 = 0 (E is indiscernible)

(x + E)^2 = x^2 + Ex^2'

x^2 + 2Ex + E^2 = x^2 + Ex^2'

2Ex = Ex^2'

x^2' = 2x

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