Graph and derivative of y=x

Question

I'm having trouble understanding why the graph of $y=x$ is different from the graph of $y = \sqrt{ x^2 }$. Aren't both equations the same once you simplify the second one? And isn't the derivative of each the same?

Answer 1

It's different. $$y=\sqrt{x^2}=|x|$$ that is different from $y=x$ The derivate of $y=x$ is $y=1$ while the derivate of $y=|x|$ is $y=1 $ for $x>0$ and $y=-1$ for $x<0$

Answer 2

$(x^2)^{1/2}=|x|$.

For positive $x$, $(x^2)^{1/2}=x$, and $(-x)^2=x^2$ too, so it gives $x$ for negative numbers too, but $y=x$ gives negative number for negative $x$.

$\frac{dx}{dx}=1$

$\frac{d(x^2)^{1/2}}{dx}, u=x^2$, then the derivate is: $\frac{du^{1/2}}{du}*\frac{du}{dx}=\frac{1}{2u^{1/2}}*2x=\frac{2x}{2(x^2)^{1/2}}=\frac{x}{(x^2)^{1/2}}=sgn(x)$

Answer 3

Consider $x=-1$:

$ \begin{align} x&=-1 \\ \sqrt{x^2}&=\sqrt{(-1)^2} =\sqrt{1}=\pm1 \end{align} $

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Every root has two solutions: $\sqrt{a}=\pm a$, for example $2 \cdot 2 = -2 \cdot -2 = 4$

Because a function can only have one $y$ for every $x$, one solution (the negative) has to be omitted. This causes the graph of $y=\sqrt{x^2}$ to always be positive.

If you would plot a graph with both the positive and negative solution of $y=\sqrt{x^2}$, it would look like $y=x$ and $y=-x$ united.

Answer 4

If $0\le a$ then $\sqrt{a}$ is defined to be that non-negative number which has square equal to $a$, so $\sqrt{a}\ge 0$ no matter what happens. In the same time $\sqrt{a^2}$ is defined to be $\vert{a}\vert$ which is defined to be the distance of $a$ from $0$ on the numberline and as such has the following properties $$ \vert{a}\vert =a\quad if \quad a\ge 0 $$ $$ \vert{a}\vert=-a\quad if \quad a<0 $$ So when in an equation $$ x^2=4 $$ you take square root on both sides you end up with $$ \vert x\vert=2 $$ and there are two numbers that have distance two from $0$. So you get $$ x=\pm 2. $$ The OP has alredy accepted an answer, I just tried to give a more "algebraic" point of view to the mix.