How is this expression simplified?

I have this:

$$\sqrt{(dx)^2 + (dy)^2}$$

And my book simplified it as:

$$\sqrt{1 + \Big(\frac{dy}{dx}\Big)^2} \times dx$$

I don't have even a close idea how he did it. If it helps, is about path lenght whit integration.

-
Your last differential in the second part should be outside the radical, not inside it. – Arturo Magidin Jun 22 '12 at 18:21

$$a\sqrt{r} = \sqrt{a^2(r)}\quad\text{if }a\gt 0\text{ and } r\gt 0.$$ So, using changes instead of differentials: \begin{align*} \sqrt{1 + \left(\frac{\Delta y}{\Delta x}\right)^2} \Delta x &= \sqrt{\left(\Delta x\right)^2\left(1 + \left(\frac{\Delta y}{\Delta x}\right)^2\right)}\\ &= \sqrt{(\Delta x)^2 + (\Delta y)^2}. \end{align*} Taking limits as $\Delta x\to 0$ converts $\Delta x$ to $dx$, $\Delta y$ to $dy$, and $\frac{\Delta x}{\Delta y}$ to the derivative $\frac{dy}{dx}$.
@Andres: The first rule is just a consequence of two things: (i) if $a$ and $b$ are both positive, then $\sqrt{ab} = \sqrt{a}\sqrt{b}$; and (ii) $\sqrt{r^2} = |r|$. Putting them both together, if $a$ and $b$ are positive, then $a\sqrt{b} =$\sqrt{a^2}\sqrt{b}=\sqrt{a^2b}$. – Arturo Magidin Jun 22 '12 at 18:32 Thank you :) Where you learn that kind of things ? I am at first year in University. Not EEUU, Argentina :) – Andres Jun 22 '12 at 18:39 @Andres This is standard root manipulation, which I believe is taught in second or third year in high school in Argentina. – talmid Jun 22 '12 at 18:41 @Andres I suggest khanacademy.org/math/algebra/exponents-radicals, for example. Or pick any third-year high school math book. Properties and tricks of algebraic manipulation like these are ubiquitous in higher-level courses, and people often use them without pointing out what they are doing, like in this case. – talmid Jun 22 '12 at 18:55$\displaystyle \sqrt{(dx)^2 + (dy)^2} = \sqrt{\left(1 + \frac{(dy)^2}{(dx)^2}\right)\cdot(dx)^2} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \cdot \sqrt{(dx)^2}= \sqrt{1 + \left(\frac{dy}{dx}\right)^2}\cdot dx$(Note that we're treating$dx$and$dy$as numbers, but that's another issue; see for example this question.) - It helps if you think about the Pythagorean theorem, first, noting how a right triangle (with legs$dx$and$dy$) has its hypoteneuse be$\sqrt{(dx)^{2}+(dy)^{2}}$; we want to change this expression into a differential, i.e., something that looks like$f(x)\,dx$. We can do this using the algebraic manipulations shown in this answer... – Alex Nelson Jun 22 '12 at 18:25 Thank you talmid :) Like I said in the other answer i didn't know the rule: $$a \times \sqrt{r+s} = \sqrt{a^2 \times (r+s)}$$ – Andres Jun 22 '12 at 18:36 @Andres Yes, because (assuming$a \geq 0$)$\sqrt{a^2(r+s)} = \sqrt{a^2}\sqrt{r+s} = a \sqrt{r+s}\$. – talmid Jun 22 '12 at 18:39