derivative notations d/dx, dy/dx, and d^2y/dx^2

We will discuss the derivative notations. I find it really helps to explain to calculus 1 students the difference between the notations d/dx, dy/dx, and also d^2y/dx^2 especially when they are learning derivatives. Of course, there's also the "prime notation", where we will write d/dx(x^3) as (x^3)' and get 3x^2 for the answer.

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[gadxxxx]

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.

[fly]

Your way with words made this easy to understand. You are an excellent educator. Thank you.
Happy belated Pi day

[SysReset]

You explain so good that even begin a 9th grader I can understand it very well!

[pneujai]

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

[EngMorvan]

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

[viswanathanm]

Very nice explanation, understood clearly.

[florentinosanchez]

Best video ever, thank you SO MUCH, YOU ARE THE BEST

[kiyomi_k]

this is so useful especially engeneering math. thank you god

[treya]

very helpful... i was confused by the second derivative notion in my problem set and thought it meant something like (dy)2/d(x2)

[ellekivung]

Thank you so much this video really helps me to understand derivative in calculus.

[mathmancalc]

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'?

[ictfan]

I m impressed that he is using 2 markers at a single time one handedly 😶‍🌫

[davidvolland]

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?

[alocin]

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.

[noahali-origamiandmore]

The thumbnail for the second derivative only shows d^2y/dx, not d^2y/dx^2. It bothers me.

[flamewings]

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 :>

[DAA]

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?

[fromblonmenchaves]

On a side note, why do we express df(x)/dx as f'(x). Is this an arbitrary decision or is there more reason?

[mohakrana]

sir please reply if we have to find d^2y / d^2x we have to multiply both first and second derivative ??

[allineediscake]

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.