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.

  • 4
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.