Calculus Calculus Math Forum

 November 8th, 2018, 07:37 AM #1 Newbie   Joined: Jun 2018 From: Jordan Posts: 24 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: 7,697 Thanks: 2681 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: 24 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^(n-1)]/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: 1,310
Thanks: 552

Quote:
 Originally Posted by Integraluser 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^(n-1)]/math] for example), but it is defined as a real number if we calculate it using the formal definition.
First, the formal definition is what ultimately counts. But you still have not given an example of where the "rules" of differentiation purportedly contradict the formal definition. Are you asserting that

$n \in \mathbb Z^+ \text { and } f(x) = x^n \not \implies f'(x) = nx^{(n-1)}.$

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: 24
Thanks: 0

Quote:
 Originally Posted by JeffM1 First, the formal definition is what ultimately counts. But you still have not given an example of where the "rules" of differentiation purportedly contradict the formal definition. Are you asserting that $n \in \mathbb Z^+ \text { and } f(x) = x^n \not \implies f'(x) = nx^{(n-1)}.$ 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.
Nah, but he asked about what rules did I mean but apparently he meant which rules do have flaws. I'll talk about what I've experienced with product rule (f*g)'(x) = f(x)g'(x) + g(x)f'(x).

If we were to find the derivative of the function $\displaystyle f(x) = \large\sqrt{x} \sin(x)$ at x = 0 using formal definition method:

$\displaystyle \lim\limits_{x \to 0} \frac{\sin(x)\sqrt{x}}{x}$

$\displaystyle \lim\limits_{x \to 0} \frac{\sin(x)}{x} * \lim\limits_{x \to 0} \sqrt{x}$

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:

$\displaystyle f'(x) = \cos(x) \sqrt{x} + \sin(x) \frac{1}{3\sqrt{x^2}}$

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 non-existing 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: 7,697
Thanks: 2681

Math Focus: Mainly analysis and algebra
Quote:
 Originally Posted by Integraluser Do you mean that he indirectly stated the formal definition form then replaced the numerator by the given expression?
I don't know about "indirectly".$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$ is a limit expression and the derivative is defined as that.

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: 1,310
Thanks: 552

Quote:
 Originally Posted by Integraluser Nah, but he asked about what rules did I mean but apparently he meant which rules do have flaws. I'll talk about what I've experienced with product rule (f*g)'(x) = f(x)g'(x) + g(x)f'(x). If we were to find the derivative of the function $\displaystyle f(x) = \large\sqrt{x} \sin(x)$ at x = 0 using formal definition method: $\displaystyle \lim\limits_{x \to 0} \frac{\sin(x)\sqrt{x}}{x}$ $\displaystyle \lim\limits_{x \to 0} \frac{\sin(x)}{x} * \lim\limits_{x \to 0} \sqrt{x}$ which is equivalent to 0.
But you have not given the formal definition of a derivative. You are not even close.

$x \ne 0 \text { and } f(x) = \dfrac{\sin(x)\sqrt{x}}{x} \implies f'(x) = \displaystyle \lim_{h \rightarrow 0} \left ( \dfrac{\dfrac{\sin(x + h)\sqrt{x + h}}{x + h} - \dfrac{\sin(x)\sqrt{x}}{x}}{h} \right ).$

I'll let you do the algebra on that.

Quote:
 Originally Posted by Integraluser 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:
We are certainly NOT interested in f'(0) because f(x) is not continuous at x = 0 and so is not differentiable there. We are in fact trying to find the formula for f'(x) at every point other than x = 0. NOTICE that, in the formal definition, h is what must approach zero, not x.

Quote:
 Originally Posted by Integraluser 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 non-existing 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.
Yes, it is a process. A very important law of limits, which I am not going to prove right now, is this

$a = b \text { and } \displaystyle \lim_{a \rightarrow c} = d \in \mathbb R \implies \lim_{b \rightarrow c} = d.$

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 non-standard 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: 683
Thanks: 456

Math Focus: Dynamical systems, analytic function theory, numerics
Quote:
 Originally Posted by JeffM1 We are certainly NOT interested in f'(0) because f(x) is not continuous at x = 0 and so is not differentiable there. We are in fact trying to find the formula for f'(x) at every point other than x = 0. NOTICE that, in the formal definition, h is what must approach zero, not x.
$f$ is continuous at 0. However, you are correct that it isn't differentiable at 0. Moreover, the product rule states something like "Suppose $f = gh$ and $g,h$ are differentiable....."

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: 7,697 Thanks: 2681 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: 1,310
Thanks: 552

Quote:
 Originally Posted by SDK $f$ is continuous at 0. However, you are correct that it isn't differentiable at 0. Moreover, the product rule states something like "Suppose $f = gh$ and $g,h$ are differentiable....." 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.
SDK

I was discussing his later example of $\dfrac{\sin(x)\sqrt{x}}{x}.$

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. Tags derivatives, limits, questions Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post jkh1919 Calculus 9 October 6th, 2012 04:21 PM r-soy Calculus 3 November 28th, 2010 12:31 PM r-soy Calculus 3 November 4th, 2010 03:42 PM r-soy Calculus 2 November 3rd, 2010 11:31 PM r-soy Calculus 2 November 3rd, 2010 10:25 PM

 Contact - Home - Forums - Cryptocurrency Forum - Top      