March 18th, 2019, 05:35 PM  #1 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Is dx the limit of Δx?
So, I'm trying to deeply understand some things that I take for granted in mathematics. Let's start with some displacements. Given two points A(x, f(x)) and B(x0, f(x0)) $\displaystyle \frac{\Delta{f}}{\Delta{x}} = \frac{f(x_0)  f(x)}{x_0  x} $ Lets name $\displaystyle h = x_0  x $ Solving for x0 $\displaystyle x_0 = x + h $ Substituting to the original displacement ration $\displaystyle \frac{\Delta{f}}{\Delta{x}} = \frac{f(x + h)  f(x)}{h} $ Taking the limit on both sides such us h>0 $\displaystyle \lim_{h\to0}\frac{\Delta{f}}{\Delta{x}} = \lim_{h\to0}\frac{f(x + h)  f(x)}{h} \Leftrightarrow \frac{\lim_{h\to0}\Delta{f}}{\lim_{h\to0}\Delta{x} } = \lim_{h\to0}\frac{f(x + h)  f(x)}{h} \Leftrightarrow \frac{df}{dx} = \lim_{h\to0}\frac{f(x + h)  f(x)}{h} $ Question 1 So in other words dx is still a length like Δx but dx <<<<<< Δx in such a way that dx is as small as possible. So is the below an accepted writing for dx? $\displaystyle \lim_{x_0x\to0}\Delta{x} = dx$ And $\displaystyle \lim_{x_0x\to0}\Delta{f} = df$ Question 2 Is this why we can rearrange equations like this? (We treat dx, df as limit values) $\displaystyle v = \frac{dx}{dt} \Leftrightarrow dx = v\cdot dt $ Question 3. If Question 2 is true then it means I can cancel any df dx? $\displaystyle v = \frac{dx}{dt} \Leftrightarrow dt \cdot v = \frac{dx}{dt} \cdot dt $ Question 4: Can I treat the symbol of integration exactly like I do with numbers when I multiply or divide both sides of an equation? $\displaystyle a = \frac{dv}{dt} \Leftrightarrow dv = a \cdot dt $ Multiplying both sides with integration??? lol but I know this can happen $\displaystyle \int dv = \int a \cdot dt $ Question 5: When I have a differential equation I can multiply both sides with a dx and an integration to complete the form of integration? $\displaystyle \frac{dy}{dx} = yx \Leftrightarrow \frac{dy}{dx} \cdot \frac{1}{y} = x $ Multiplying both sides with dx (dx's on the left cancel out) $\displaystyle \frac{dy}{dx} \cdot \frac{1}{y} \cdot dx = x \cdot dx \Leftrightarrow \frac{1}{y} \cdot dy = x \cdot dx $ Integrating both sides: $\displaystyle \int \frac{1}{y} \cdot dy = \int x \cdot dx $ So in general i can do whatever I want with dx's, df's and integrals in both sides of an equation like when I do with numbers or variables? Well the truth is that I'm using all these stuff but nobody told me that. By seeing how our professors are teaching different stuff in mathematics, I just noticed that these operations are actually matching the same techniques we use in algebra so I decided to treat them like that. I hope I'm doing things right. Thank you 
March 19th, 2019, 03:19 AM  #2 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 
$\displaystyle dx$ is the limit of delta $\displaystyle x$. For the integration part , you must use the word ”integration of equality” or ”add both sides the integral” and not multiply . Example : Since $\displaystyle e^x >x \; \Rightarrow \int_{D} e^x dx > \int_{D} xdx $. Let $\displaystyle D\in [0,1] $ , $\displaystyle e^x _{0}^{1} > 1/2 x^2 _{0}^{1}$. Which gives $\displaystyle e1>1/2$ without calculating them . Last edited by idontknow; March 19th, 2019 at 03:22 AM. 
March 19th, 2019, 04:46 AM  #3 
Senior Member Joined: Dec 2015 From: somewhere Posts: 605 Thanks: 88 
About the differential equation : $\displaystyle y’=yx$, you lose some solutions if you divide by y. The general solution is found with the integrating factor which is $\displaystyle z(x) $ where $\displaystyle \frac{z’ }{z} = x$ . When you substitute $\displaystyle x$ to the equation the left side of equation is $\displaystyle zy’+yz’=(zy)’$ and in the right side of equation you have an easy integral . 
March 19th, 2019, 06:41 AM  #4 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
It's complicated. Especially with regard to your third question. Differentials are things, but they aren't limits of $\Delta x$ because the limit value of $\Delta x$ is zero. Also, the symbol $\frac{\mathrm df}{\mathrm dx}$ is not a ratio of differentials, it is an operator $\frac{\mathrm d}{\mathrm dx}$ acting on the function $f$. I tend to think of differentials as infinitesimals after the manner in which Newton used $\epsilon$ and conversion between those and derivatives as just something that works. Wiki page on differentials 
March 19th, 2019, 06:53 AM  #5  
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8  Quote:
 
March 19th, 2019, 08:48 AM  #6 
Senior Member Joined: Oct 2015 From: Greece Posts: 137 Thanks: 8 
Also what If I do this: Let's say I have the function $\displaystyle f(x) = x $ then $\displaystyle \frac{df}{dx} = 1 $ Is the below wrong? $\displaystyle df = dx \Leftrightarrow df \cdot dx = dx^2 \Leftrightarrow \int df \cdot dx = \int dx^2 $ 
March 19th, 2019, 11:58 AM  #7 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra  It works because $$\mathrm dy = \frac{ \mathrm dy }{ \mathrm dx } \mathrm dx$$

March 20th, 2019, 08:35 AM  #8 
Member Joined: Feb 2019 From: United Kingdom Posts: 44 Thanks: 3 
The most embarrassing thing about this post is not on the OP but the glorified “Math Team” who don’t give a sh*t. Who can spot what’s wrong with the OP’s definition of derivative? How on Earth that slipped past by this so called “team” I don’t know. Either give up your gloves for the sake of your reputation or be active to correct people.

March 20th, 2019, 08:58 AM  #9 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,675 Thanks: 2655 Math Focus: Mainly analysis and algebra 
I'm comfortable with all the answers I gave. I never claimed to be correcting every error (that isn't my job) nor did I even claim to be answering every question. At least I was able, in the short time available to me, to contribute something positive and useful to the thread.

March 20th, 2019, 09:29 AM  #10  
Math Team Joined: May 2013 From: The Astral plane Posts: 2,258 Thanks: 929 Math Focus: Wibbly wobbly timeywimey stuff.  Quote:
Personally I feel that it's been fixed "off to the side." As $\displaystyle \lim_{h \to 0} \Delta x = dx$ has been pointed out to be inaccurate I, personally, feel the problem has been resolved. But if you disagree then point it out. Dan  

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