Derivative equals reciprocal

In this video, as a sequel to my video on f’ = f^-1, I find all the functions whose derivative equals to its reciprocal. As an added bonus, I also find all the functions whose derivative gives the square of the function. Enjoy!

[RyanLucroy]

Well, I used a different approach, since they are both seperable, it's pretty easy to do:

y' = 1/y
dy/dx = 1/y | * y * dx
y*dy = dx | Integrate
1/2 * y^2 = x + c | Solve for y
y = +- sqrt (2x+c)

And the same goes with y'=y^2

[camdamcool]

How about combine the three... Derivative of a function equals one over the square of the inverse :D

[BALAGE]

Not even half as nice as when the derivative equals the inverse.

[qarsiseer]

Not really familiar with the Chen Lu-ing(?) kind of confuses me. Using Liebnitz and separation of variables is far easier for me intiutively but I sorta get it? Maybe there’s context that’s missing that would help?

[someone]

Finally a doctor Peyam's video that I can understand

[Royvan]

phrasing it as a reverse chain rule is an interesting way to look at it. my first instinct would be that it's separable and separate it into two indefinite integrals of dy and dx.

[geekjokes]

GENERALIZE ALL THE THINGS
show the general solution to a differencial equation of y'=y^p for any p (if its possible)

[hongqinzou]

Are there ones satisfy y'=y(y(x))? Clearly zero is a solution.

[philipphoehn]

Do derivative equals function composed with itself.

[PackSciences]

At 4:23, you said that we have to worry about the uniqueness about the solution.
Well y(t) = 0 is the trivial solution, then y(t) = -1/(x+c).
So we can ask ourselves, "can we build a continuous function that satisfies the diff equ on R?"
Well we need it to be differentiable at all points of the interval, so at least continuous.
Then, we need to glue the two intervals together, so that f(a+) = f(a-).
However, we notice that it's not possible on the reals to have a "gluing" of diff equ, as y_1 = 0 and y_2 = -1/(x+c), and -1/(x+c) doesn't equal zero in reals (only in extended real line).
Therefore, I think the solutions are:
- y(t) = 0
- -1/(x+c) = 0
And that's all (no piecewise-defined differentiable function).
However, you said that we have to be careful at 4:26, so did I miss something/ a solution?

[helloitsme]

You have a really creative way of solving these pde's

[ThePianofreaky]

Now find a function where the derivative is the function applied 2 times. f'(x) = f(f(x))