When I learn't calculus at school in Australia, we were taught to pronounce the second derivative as "d two y dx squared". That makes better sense now.
When it was 1859, the first Chinese calculus book "代微積拾級" was written, the notation was like d(x²)=2xdx 彳(天^二)=二天彳天 but similar notation as nowadays for dy/dx and second derivative u can find the book in harvard library
When using the above “Leibniz Notation”, “d” cannot represent a variable or a constant, because the “d” is a symbol representing the Derivative. Capital “D” represents taking the derivative of an equation, e.g. D[x^2+3x+2=0] is 2x+3=0
D [ ] is same as d/dx( ) D[y=x^2+3x+2] Taking the derivative of a function. dy=2xdx+3dx+0 Factor out the dx. dy=(2x+3)dx Divide both sides by dx. dy/dx= 2x+3
just to clear out the answer to that. Let's suppose f(x,d) is the function with variables x and d. Then the notation for the second derivative with respect to d would be: d^2f/dd^2 (or if we set y=f(d,x), we would get d^2y/dd^2) If this is allowed notationwise, please rename your variables so that d isn't a variable when taking the d:th derivative. it would be much clearer what you do if you just take the time to rename d into anything else. That would be even if you rename it into just a smiley.
Non math major here with a question about what d/dx(dy/dx) says. d/dx is a operator “take derivative of” and dy/dx says “the derivative is = to something.” So d/dx(dy/dx) says take the derivative of a derivative so the derivative of a derivative is the second derivative denoted, d/dx(dy/dx) and it equals d^2y/dx^2, and this expression reads the second derivative of of y with respect to the variable x a second time. Correct ? So when he say “we more eloquently say dx^2 not not (dx)^2 because (dx)^ 2 is treating this like it multiplying and obscures the fact we are taking a second derivative, correct?
Do we not use () for dx^2 because dx is not d*x but a single variable itself? So it would be the same as (dx)^2 but is unnecessary like how we don't use (2)^2 for 2^2. I think I understand my problem now.
f'(x) (in Lagrange's notation) is equivalent to df/dx in Leibniz's notation. f' is probably used because it is faster to write than df/dx but they actually mean the same. Lagrange's notation has a slight benefit over df/dx when it comes to evaluating the derivative. For example, let's say you want to evaluate the derivative of f(x) at x=0. In Leibniz's notation this would be slightly complicated because you have to write df/dx|x=0. Or (with Lagrange's) you can just write f'(0).
The second derivative tells us about the concavity of the curve. If the second derivative is positive over a certain interval, the graph is concave up at that interval. The converse is also true.
When I saw this firstly, I had only one fact: d/dx - it’s just derivative notation: dy/dx = y'. And after a few text I saw d2y/dx^2. And guess what? I just took d(dy/dx)/dx. And I thought: okey, when we take derivative, we do derivative of function w.r.t. x, so we need not Cousin rule, we need twice do derivative and do it twice w.r.t x. But… dx^2 just for beauty :>
I've seen in textbooks people multiplying in both sides by dx. Does that makes sense? Is it an entity or an operation? Which and when algebra rules apply?
I think in the US and may be in other countries from the English speaking world it is very common. I studied physics in Germany many many years ago, and this kind of operations my maths professors and lecturers of maths wouldn't have been amused of.
From what my teacher explained to me, what dx actually represents is an infinitesimal change in the x value. Usually I only see people doing that in differential equations, and sometimes integrals
Yes. An integral in the x world is the sum (which is why it's an s) of your y values (the height of the rectangles) times dx (the infinitesimal width). d/dx is an operation sort of. The dx or dy itself is a differential that you can multiply or divide by to change the form to something integrable, for example.
The Leibniz notation is very intuitive and convenient in many cases but I have to say I dislike it a lot for the following reason (among others): Let's define the functions f:R->R, f(x)=x^2 and g:R->R, g(t)=t^2. f and g have the same domain and codomain and they both map an element of the domain to the codomain by squaring it. Therefore, f and g are exactly the same, I just chose to call the elements of the domains differently when I defined the functions. Obviously, f and g will also have the same derivative, but the Leibniz notation might mislead us into writing df/dx=2x and dg/dx=0 since, as it's often said, "g does not depend on x". If d/dx is supposed to stand for the "operator" that maps a function to its right to its derivative that should work no matter what I called the input of the function. The point I want to make is the following: The concept of the derivative of a function exists without the need to call the input of that function a certain name and therefore the notation of the derivative should not refer to that arbitrarily chosen name of the function's input when it was defined. The letter x simply has no meaning outside the definition of the function. This is not criticizing blackpenredpen by the way. The notation is in common use and it does have its advantages, so it would be wrong not to explain how it works.
I don’t think it’s a big problem. Differentiation as an operation is always with respect to a variable so it is okay that the operator notation needs to specify the variable. Lagrange and Newton notation can be ambiguous if the independent variable is ambiguous.
If your acceleration changes depending on the time you've travelled, it won't change with respect to the friction of the floor unless you can parameterise both under am equal unit. If g=g(t), then dg/dx is equal to 0 as x and t represent different things. Your point sort of makes sense, but when using dummy variables, you can't pick and mix them as you please without telling us how they relate to each other. If you said let t=x or something, you'd be justified in saying dg/dx ≠ 0 but as long as they are unrelated variables to our knowledge, then it will differentiate to 0
A simple way to think of it: d/dx literally says find the derivative of a function with respect to “x”: d/dx [y=x^2 +3x+2] is dy=2xdx+3dx which is dy/dx=2x+3 after both sides are divided by dx.
At 3 minutes 27 seconds. Why is it bad to write (dx)^2 to me this looks more intuitive where as the formal way of dx^2 makes it look like only the x is being squared and not the d.
You forgot to explain each d/dx, dy/dx, d^2y/dx^2 etc represents a Rate, not a fraction, but a division expressed in Rate form where the “d”s don’t cancel each other out, because “d” is a symbol and not a variable or constant, but a symbol representing the term “derivative”. lol. A capital “D” can be used to represent taking the Derivative of an equation. e,g, D[ x^2+3x+2=0] is 2x+3=0.
Your mistake is differentiating with respect to a constant. That just doesn't make any sense The derivative of a function is a rate of change of that function (technically it's the limit of that rate of change between two points as the difference between those points approaches 0, but whatever). If x is changing, so is f(x), and the derivative is just a very-very-very small change in f(x) divided by a very-very-very small change in x. That's what df/dx says - df and dx are just very-very-very small changes. Now, if x is not changing, that means a small change in it just never happens. It's 0, pretty much. So when you're differentiating with respect to a constant, you're dividing by 0, essentially. Not 'zero plus', not 'zero minus', just the 0 we're all used to. There's no change, even a small one. And ya know division by zero results in some wild stuff. Always. Not very rigorous, but you know what I mean
Couldn't d/dx be like D sub x? I.e. capital D and a sub small x. And what about dy/dx ; wouldn't that be like d/dx (y) as well as D sub x of y or Dx y. d2y/dx2 would be Dx(Dx y) , or D2x y , or best Dxx y. Isn't it all basically abuse of notation?
When I learn't calculus at school in Australia, we were taught to pronounce the second derivative as "d two y dx squared". That makes better sense now.
he didn't learn it, he learn't it
@@hellohabibi1 learnt?
@@gadxxxx you used learn't in your original comment, they're mocking you
Your way with words made this easy to understand. You are an excellent educator. Thank you.
Happy belated Pi day
or Pi +1 if you prefer :p
Thanks.
You explain so good that even begin a 9th grader I can understand it very well!
When it was 1859, the first Chinese calculus book "代微積拾級" was written, the notation was like
d(x²)=2xdx
彳(天^二)=二天彳天
but similar notation as nowadays for dy/dx and second derivative
u can find the book in harvard library
That is finding differential d(x²) 微分 , differentiate x² without repect to anything . Term dy/dx is derivative 导数, differentiate y wrt to x.
this is so useful especially engeneering math. thank you god
Also, there is dy/dx², the derivative of y with respect to x². 😁
E.g.: y=x⁴
dy/dx² = dx⁴/dx² = d(x²)²/dx² = du²/du = 2u = 2x²
Best video ever, thank you SO MUCH, YOU ARE THE BEST
Very nice explanation, understood clearly.
So just to test the students, what is the notation for the second derivative with respect to 'd' of a function with variables 'x' and 'd'?
d2x/dd2 I think?
When using the above “Leibniz Notation”, “d” cannot represent a variable or a constant, because the “d” is a symbol representing the Derivative. Capital “D” represents taking the derivative of an equation, e.g. D[x^2+3x+2=0] is 2x+3=0
D [ ] is same as d/dx( )
D[y=x^2+3x+2] Taking the derivative of a function.
dy=2xdx+3dx+0 Factor out the dx.
dy=(2x+3)dx Divide both sides by dx.
dy/dx= 2x+3
just to clear out the answer to that.
Let's suppose f(x,d) is the function with variables x and d. Then the notation for the second derivative with respect to d would be:
d^2f/dd^2
(or if we set y=f(d,x), we would get d^2y/dd^2)
If this is allowed notationwise, please rename your variables so that d isn't a variable when taking the d:th derivative. it would be much clearer what you do if you just take the time to rename d into anything else. That would be even if you rename it into just a smiley.
@@lajont d:th derivative? Do you mean nth derivative by that or what?
I m impressed that he is using 2 markers at a single time one handedly 😶🌫
right before you said "dont put the parenthesis on dx squared" i put the parenthesis thinking i was a genius 😂 anyways nice video!
very helpful... i was confused by the second derivative notion in my problem set and thought it meant something like (dy)2/d(x2)
*notation
Thank you so much this video really helps me to understand derivative in calculus.
I saw the pokemon ball and wasted 3 hours watching pokemon 💀
Non math major here with a question about what d/dx(dy/dx) says. d/dx is a operator “take derivative of” and dy/dx says “the derivative is = to something.” So d/dx(dy/dx) says take the derivative of a derivative so the derivative of a derivative is the second derivative denoted, d/dx(dy/dx) and it equals d^2y/dx^2, and this expression reads the second derivative of of y with respect to the variable x a second time. Correct ? So when he say “we more eloquently say dx^2 not not (dx)^2 because (dx)^ 2 is treating this like it multiplying and obscures the fact we are taking a second derivative, correct?
Lovely lesson, Serously helped
i'm just here giggling at him carrying the pokeball plush😂 so random
Why we don't use parenthesis
The thumbnail for the second derivative only shows d^2y/dx, not d^2y/dx^2. It bothers me.
My bad. I will change it when I get my iPad.
Thanks you for this content.
is there a way to notate the second derivative as a verb?
d^2/dx^2
Edit: I know understand, dx is not d*x but a single object itself.
does the dx^2 notation means (dx)^2 but is just prefered to be written so?
Do we not use () for dx^2 because dx is not d*x but a single variable itself? So it would be the same as (dx)^2 but is unnecessary like how we don't use (2)^2 for 2^2. I think I understand my problem now.
On a side note, why do we express df(x)/dx as f'(x). Is this an arbitrary decision or is there more reason?
f'(x) (in Lagrange's notation) is equivalent to df/dx in Leibniz's notation. f' is probably used because it is faster to write than df/dx but they actually mean the same.
Lagrange's notation has a slight benefit over df/dx when it comes to evaluating the derivative. For example, let's say you want to evaluate the derivative of f(x) at x=0. In Leibniz's notation this would be slightly complicated because you have to write df/dx|x=0. Or (with Lagrange's) you can just write f'(0).
sir please reply if we have to find d^2y / d^2x we have to multiply both first and second derivative ??
Thank you teacher
Sensei*
Why i cant make it d2x /d2x2 ?
Thank you professor, but your audio has a lot of echo and very difficult to understand what you are saying. But thank you for clearing the concept.
*d²/dx² = d²/(dx)²* ?
Why we can't write d²y/ d²x² or d²y/d²x
Big thanks!
Awesome thanks for the great explanation! :)
Can you explain me the clear purpose of finding the second derivative and what does it really represent in a curve?
You can use it to find the largest slope of a curve, or you can use it to find “inflection” points
The second derivative tells us about the concavity of the curve. If the second derivative is positive over a certain interval, the graph is concave up at that interval. The converse is also true.
The second derivitave tells us about how the graph is curving or the amount of curvature.
The curve’s Maxima or Minima, positive slope (rate), negative slop (rate), point of inflection.
@@mikeb6389 do you mean like taking the 1st derivative and setting it equal to zero to find the maximum’s and minimum‘s of the function?
When I saw this firstly, I had only one fact: d/dx - it’s just derivative notation: dy/dx = y'. And after a few text I saw d2y/dx^2. And guess what? I just took d(dy/dx)/dx. And I thought: okey, when we take derivative, we do derivative of function w.r.t. x, so we need not Cousin rule, we need twice do derivative and do it twice w.r.t x. But… dx^2 just for beauty :>
I've seen in textbooks people multiplying in both sides by dx. Does that makes sense? Is it an entity or an operation? Which and when algebra rules apply?
I think in the US and may be in other countries from the English speaking world it is very common. I studied physics in Germany many many years ago, and this kind of operations my maths professors and lecturers of maths wouldn't have been amused of.
From what my teacher explained to me, what dx actually represents is an infinitesimal change in the x value. Usually I only see people doing that in differential equations, and sometimes integrals
Yes. An integral in the x world is the sum (which is why it's an s) of your y values (the height of the rectangles) times dx (the infinitesimal width).
d/dx is an operation sort of. The dx or dy itself is a differential that you can multiply or divide by to change the form to something integrable, for example.
@@hassanalihusseini1717 as my pphysics professor said: "you cannot multiply parts of an operator separately - but in this case it works, so we do it"
@@Engy_Wuck ThT's the right approach!
i really want to enter the stage where i can start expecting the answers but idk how to help myself learning this
wait maybe i should pretend knowing the answer after i watched the video
The Leibniz notation is very intuitive and convenient in many cases but I have to say I dislike it a lot for the following reason (among others): Let's define the functions f:R->R, f(x)=x^2 and g:R->R, g(t)=t^2. f and g have the same domain and codomain and they both map an element of the domain to the codomain by squaring it. Therefore, f and g are exactly the same, I just chose to call the elements of the domains differently when I defined the functions. Obviously, f and g will also have the same derivative, but the Leibniz notation might mislead us into writing df/dx=2x and dg/dx=0 since, as it's often said, "g does not depend on x". If d/dx is supposed to stand for the "operator" that maps a function to its right to its derivative that should work no matter what I called the input of the function.
The point I want to make is the following: The concept of the derivative of a function exists without the need to call the input of that function a certain name and therefore the notation of the derivative should not refer to that arbitrarily chosen name of the function's input when it was defined. The letter x simply has no meaning outside the definition of the function.
This is not criticizing blackpenredpen by the way. The notation is in common use and it does have its advantages, so it would be wrong not to explain how it works.
I don’t think it’s a big problem. Differentiation as an operation is always with respect to a variable so it is okay that the operator notation needs to specify the variable. Lagrange and Newton notation can be ambiguous if the independent variable is ambiguous.
If your acceleration changes depending on the time you've travelled, it won't change with respect to the friction of the floor unless you can parameterise both under am equal unit.
If g=g(t), then dg/dx is equal to 0 as x and t represent different things.
Your point sort of makes sense, but when using dummy variables, you can't pick and mix them as you please without telling us how they relate to each other. If you said let t=x or something, you'd be justified in saying dg/dx ≠ 0 but as long as they are unrelated variables to our knowledge, then it will differentiate to 0
Really nice video, but I'm not getting when I must use d/dx or dy/dx. Could you (or someone) explain it to me?
A simple way to think of it: d/dx literally says find the derivative of a function with respect to “x”: d/dx [y=x^2 +3x+2] is dy=2xdx+3dx which is dy/dx=2x+3 after both sides are divided by dx.
(Thank you good sir)
At 3 minutes 27 seconds. Why is it bad to write (dx)^2 to me this looks more intuitive where as the formal way of dx^2 makes it look like only the x is being squared and not the d.
It was very informative thanks
Btw...did your english teacher ever scold you for your 'r' and 'v' ?😂...just asking for fun
very nice
You forgot to explain each d/dx, dy/dx, d^2y/dx^2 etc represents a Rate, not a fraction, but a division expressed in Rate form where the “d”s don’t cancel each other out, because “d” is a symbol and not a variable or constant, but a symbol representing the term “derivative”. lol. A capital “D” can be used to represent taking the Derivative of an equation. e,g, D[ x^2+3x+2=0] is 2x+3=0.
X=constant
=> d/dx (x) =d/dx(constant)
=> 1 =0
??????????????
Huh?
Your mistake is differentiating with respect to a constant. That just doesn't make any sense
The derivative of a function is a rate of change of that function (technically it's the limit of that rate of change between two points as the difference between those points approaches 0, but whatever). If x is changing, so is f(x), and the derivative is just a very-very-very small change in f(x) divided by a very-very-very small change in x. That's what df/dx says - df and dx are just very-very-very small changes.
Now, if x is not changing, that means a small change in it just never happens. It's 0, pretty much.
So when you're differentiating with respect to a constant, you're dividing by 0, essentially. Not 'zero plus', not 'zero minus', just the 0 we're all used to. There's no change, even a small one.
And ya know division by zero results in some wild stuff. Always.
Not very rigorous, but you know what I mean
@@eliasmazhukin2009 thanks sir...
One of the nice thing about Learning and Understanding Calculus is no more graphing of equations.lol
Couldn't d/dx be like D sub x?
I.e. capital D and a sub small x.
And what about dy/dx ; wouldn't that be like d/dx (y) as well as D sub x of y or Dx y.
d2y/dx2 would be Dx(Dx y) , or D2x y , or best Dxx y.
Isn't it all basically abuse of notation?
Thx
Error in the thumbnail mate
just realised the pokeball is his mic
Isn’t it better to say things like “d squared y d x squared” instead of “d 2…”?
I've always heard it as "d two y over d x squared".
The Ds don’t cancel
درود به شرفت مرد !
sorcery