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This function is from my text: $$p(\theta) = \sqrt{13\theta}$$

It states that the derivative of the function $p(\theta)$ with respect to the variable $\theta$ is the function $p'$ whose value at $\theta$ is given by the following formula, provided that the limit exists:

$$\lim_{z \to \theta} \frac{p(z)-p(\theta)}{z-\theta}$$

Why is $z$ approaching $\theta$? I have always seen the definition of the derivative as the limit of a function as the increment $h \rightarrow 0$:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Why can't we say: $\sqrt{13(\theta+z)}$ as $z \rightarrow 0$? I don't understand the limit derivative definition being applied to this problem by having $z \rightarrow \theta$.

Thank you.

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    $\begingroup$ It's the same thing. Use the change of variable: $z=x+h$. $\endgroup$ – Git Gud Nov 17 '13 at 23:30
  • $\begingroup$ Please, people, don't let my comment act as answer. Someone please give a decent answer. $\endgroup$ – Git Gud Nov 17 '13 at 23:40
  • $\begingroup$ @GitGud thanks, I'm still not understanding though. $\endgroup$ – Emi Matro Nov 17 '13 at 23:51
  • $\begingroup$ user436158 I've added an answer going through the steps which Git Gud's comment mentions, hopefully explaining the two definitions are the same. $\endgroup$ – coffeemath Nov 18 '13 at 0:17
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You can use either of the definitions $$p'(\theta)=\lim_{h \to 0}\frac{p(\theta+h)-p(\theta)}{h}, \tag{1}$$ which seems close to your approach, or else the one suggested in the text, $$p'(\theta)=\lim_{z \to \theta}\frac{p(z)-p(\theta)}{z-\theta}. \tag{2}$$ For this problem, definition $(2)$ might look cleaner in applying it, since the technique (in either case) is to multiply top and bottom by the conjugate of the difference of radicals, and definition $(2)$ makes the radicals simpler to manipulate.

Nevertheless the definition $(1)$ has the advantage that, no matter what the algebra is, one is always looking to at some point factor out $h$ from something and cancel it, so that at that point making $h \to 0$ will give an answer using limit laws.

The reason these are both the same is that, if you believe $(1)$, then defining $h=z-\theta$ we see that $h \to 0$ is the same as $z \to \theta$, and also $\theta +h$ becomes replaced by $z$. Thus everything matches up in the two definitions.

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