November 8th, 2018, 07:37 AM  #1 
Newbie Joined: Jun 2018 From: Jordan Posts: 19 Thanks: 0  Questions about derivatives
Hey. So we had this question: Given that $\displaystyle f(x) = f(x+h)  5x^3 h$, find $\displaystyle f'(x)$? So our teacher solved it in two ways, one of them was this: $\displaystyle f(x+h)  f(x) = 5x^3 h $ $\displaystyle \large\lim\limits_{h \to 0} \large\frac{f(x+h)  f(x)}{h} = \large\lim\limits_{h \to 0} \large\frac{5x^3 h}{h} $ $\displaystyle f'(x) = 5x^3 $ My question is regarding the second step; how was he able to insert that limit? Did he multiply by the limit of (1/h) or divide by h then taken the limit of both sides or what? My second question is, what's the difference between first derivative formal definition and derivative rules regarding the way the work? And why can derivative rules have flaws in some points meanwhile first derivative formal definition doesn't have it at the same points? Last edited by skipjack; November 9th, 2018 at 03:50 AM. 
November 8th, 2018, 09:21 AM  #2 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 
He didn't insert anything. That second step quotes the definition of the derivative. On the right hand side of the equation, he simply replaced the numerator with the equivalent expression that he had derived before. Which rules are you talking about? And what flaws? 
November 8th, 2018, 10:32 AM  #3 
Newbie Joined: Jun 2018 From: Jordan Posts: 19 Thanks: 0 
Do you mean that he indirectly stated the formal definition form then replaced the numerator by the given expression? What I mean is that, if f,g are two given functions, f has a derivative at x= a but g doesn't have a derivative at x = a, meanwhile the product function of f,g has a derivative at x=a. However, that derivative is undefined if we were to calculate it using derivative rules ([math]nx^(n1)]/math] for example), but it is defined as a real number if we calculate it using the formal definition. Last edited by skipjack; November 9th, 2018 at 03:52 AM. 
November 8th, 2018, 11:18 AM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489  Quote:
$n \in \mathbb Z^+ \text { and } f(x) = x^n \not \implies f'(x) = nx^{(n1)}.$ How do you justify that assertion? EDIT: Going back to your original post, perhaps it would have been clearer (though no more correct) had your teacher gone $f(x + h)  f(x) = 5x^3h \text { and } h \ne 0 \implies \dfrac{f(x + h)  f(x)}{h} = \dfrac{5x^3h}{h} = 5x^3$ before introducing limits. Of course, it is hard to guess what your teacher was doing when you may not have told us the original problem and when you have certainly not said what his other method was. Last edited by skipjack; November 9th, 2018 at 03:52 AM.  
November 8th, 2018, 12:19 PM  #5  
Newbie Joined: Jun 2018 From: Jordan Posts: 19 Thanks: 0  Quote:
If we were to find the derivative of the function $\displaystyle f(x) = \large\sqrt[3]{x} \sin(x) $ at x = 0 using formal definition method: $\displaystyle \lim\limits_{x \to 0} \frac{\sin(x)\sqrt[3]{x}}{x} $ $\displaystyle \lim\limits_{x \to 0} \frac{\sin(x)}{x} * \lim\limits_{x \to 0} \sqrt[3]{x} $ which is equivalent to 0. Meanwhile, if you try using derivative rules to find the derivative at zero for the product function, according to the resulted derivative function, it would be undefined at x = 0: $\displaystyle f'(x) = \cos(x) \sqrt[3]{x} + \sin(x) \frac{1}{3\sqrt[3]{x^2}} $ this is what I meant previously by flaws of derivative rules when the formal definition finds the derivative without problems. Why does this happen and is it related to the way the derivative rules work? As for the other question, that's the whole question with nothing missing, the other method doesn't matter here because it isn't related to the question (well, by introducing what does f'(x) equal in terms of limits (formal definition) and replacing $\displaystyle f(x+h)  f(x)$ with $\displaystyle 5x^3h$ nothing really special with this). And nope, I posted the steps he did exactly, without anything missing or any step forgotten, he simply moved on from the first step I posted to introducing the limit along with (1/h) both together in the same second step directly. So, did he really introduce the limit on both sides as a process (like integrating both sides for example, or taking the derivative of both sides with respect to a variable) or did he multiply both sides by a limit (like multiplying with a number, but I doubt it here though since we can't multiply something with a nonexisting limit anyways), or what exactly? That's my question, when I asked him, he just told me that he divided by h and added the limit so that h can never be equal to 0. Last edited by skipjack; November 9th, 2018 at 04:10 AM.  
November 8th, 2018, 01:29 PM  #6  
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra  Quote:
The product rule comes from manipulation of limit expressions, I hadn't noticed that inconsistency before.  
November 8th, 2018, 02:20 PM  #7  
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489  Quote:
$x \ne 0 \text { and } f(x) = \dfrac{\sin(x)\sqrt[3]{x}}{x} \implies f'(x) = \displaystyle \lim_{h \rightarrow 0} \left ( \dfrac{\dfrac{\sin(x + h)\sqrt[3]{x + h}}{x + h}  \dfrac{\sin(x)\sqrt[3]{x}}{x}}{h} \right ).$ I'll let you do the algebra on that. Quote:
Quote:
$a = b \text { and } \displaystyle \lim_{a \rightarrow c} = d \in \mathbb R \implies \lim_{b \rightarrow c} = d.$ So I said in my first post that I thought what your teacher said could have been made a bit more intuitive. He had $f(x + h)  f(x) = 5x^3h.$ Then, before fussing with limits, he could have set up and simplified the Newton quotient: $\dfrac{f(x + h)  f(x)}{h} = \dfrac{5x^3h}{h} = 5x^3 \text { if } h \ne 0.$ Now it is obvious that $\displaystyle \lim_{h \rightarrow 0} 5x^3 = 5x^3 \text { for all real } x$ because 5x^3 is not dependent on h at all. But your teacher thought that too obvious to mention. Now using that law of limits I mentioned $\dfrac{f(x + h)  f(x)}{h} = 5x^3 \implies \displaystyle \lim_{h \rightarrow 0} \dfrac{f(x + h)  f(x)}{h} = \lim_{h \rightarrow 0} 5x^3 = 5x^3.$ Wait a minute you may say. That equality obtained only when h is not equal to 0. But the limit as h approaches 0 is carefully defined to exclude h = 0. EDIT: Both standard and nonstandard analysis are very carefully developed. I personally believe that a less rigorous approach to teaching beginning calculus would make learning it a lot easier for students. Of course, the student may say, "But you haven't really proved any of this." The response that I would give is: "It is proved in analysis, but what we are teaching right now is how to use calculus, not how to prove it." Last edited by skipjack; November 9th, 2018 at 04:05 AM.  
November 8th, 2018, 02:59 PM  #8  
Senior Member Joined: Sep 2016 From: USA Posts: 503 Thanks: 281 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
This explains why the example doesn't contradict the product rule. Namely, the product rule does not give the value you claim it does because the product rule assumes both factors are differentiable. Last edited by skipjack; November 9th, 2018 at 04:25 AM.  
November 8th, 2018, 02:59 PM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,502 Thanks: 2511 Math Focus: Mainly analysis and algebra 
You aren't using the function he gave. His idea about $f'(0)$ is perfectly valid, although we'd normally have $h$ rather than $x$.

November 8th, 2018, 03:09 PM  #10  
Senior Member Joined: May 2016 From: USA Posts: 1,192 Thanks: 489  Quote:
I was discussing his later example of $\dfrac{\sin(x)\sqrt[3]{x}}{x}.$ I do not see how that function can be continuous at x = 0 when it is not even defined at x = 0. Perhaps I need to clarify what function I am talking about. Last edited by skipjack; November 9th, 2018 at 04:11 AM.  

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