Definition of a derivative - absolute value

[Dethrone]

\(\displaystyle \d{}{x}\frac{1}{\sqrt{x}}\) by the definition of the derivative.

$$\lim_{{h}\to{0}}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt{x}}}{h}=\lim_{{h}\to{0}}\frac{\sqrt{x}-\sqrt{x+h}}{h\sqrt{x^2+2xh}}=\lim_{{h}\to{0}}\frac{x-(x+h)}{h\sqrt{x^2+2xh}\left(\sqrt{x}+\sqrt{x+h}\right)}$$

Setting $h=0$:

$$=\frac{-1}{2\sqrt{x^2}(\sqrt{x})}$$

Question is why do we assume $x>0$?

EDIT: Wait. dumb question, $x$ is only defined for $x>0$ (Headbang)

 

[I like Serena]

EDIT: Wait. dumb question, $x$ is only defined for $x>0$ (Headbang)

$x$? Or \(\displaystyle \frac 1{\sqrt x}\)?

 

[Dethrone]

The latter. But I meant, the value of $x$ in the latter is defined only for $x>0$.

 

[I like Serena]

The latter. But I meant, the value of $x$ in the latter is defined only for $x>0$.

Hmm. Since $\frac{1}{\sqrt{-1}}$ is not defined in the real numbers, does that mean that $-1$ is not defined? (Thinking)

 

[Dethrone]

What I said: "The value of $x$ in the function is undefined." What I meant to say: "The function is undefined when it takes on the value of $x\le 0$." (Tmi)

 

[I like Serena]

What I said: "The value of $x$ in the function is undefined." What I meant to say: "The function is undefined when it takes on the value of $x\le 0$." (Tmi)

Hold on... (Wondering)
... if the function takes on a value (apparently $\le 0$), how can it be undefined? (Devil)

 

[Dethrone]

Hold on... (Wondering)
... if the function takes on a value (apparently $\le 0$), how can it be undefined? (Devil)

Because our original function is $\frac{1}{\sqrt{x}}$, where the function is only defined for values of $x>0$.
If the problem is because of the clarity of my phrase, I give up. (Doh)

 

[I like Serena]

Because our original function is $\frac{1}{\sqrt{x}}$, where the function is only defined for values of $x>0$.
If the problem is because of the clarity of my phrase, I give up. (Doh)

You finally got it right!
Good! (Whew)