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This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just loo

This reminds me of an exercise I'll never forget from my Math Methods course: finding the derivative of arcsin(x). It seems almost impossible because, just looking at it, there seems to be nothing you can do to simplify it. Then, out of sheer nothing-else-to-do-ism, you take the sin() of it and realize sin(arcsin(x)) = x. Take the derivative of both sides, apply chain rule and draw a right triangle and you have the answer. Like the words the author uses for the integral, it's all valuable technique.

tonyarkles 3y ago
Alternatively, x = sin(theta) and arcsin(sin(theta)) = theta, depending on the limits. It’s a cool technique, thanks for the reminder!
77pt77 3y ago
Dude. That's just a direct application of the inverse function theorem. Even geometrically you can see that swapping the axis (x and y) gives you the desired result.
- CGamesPlay 3y ago
  GP just happened to be one of the lucky 10,000 on that day in math class. https://xkcd.com/1053/ https://xkcd.com/1053/
saghm 3y ago
In turn, this reminds me a bit of a calculus problem I saw freshman year of college that I ended up sharing with my old high school calculus teacher because it similarly looks intractable until you make a "simplification". The problem was to find the integral of `x/(x + 1)` (I forget the exact bounds, presumably from 0 to x with respect to x). The trick is that this the same as `(x + 1 - 1)/(x + 1)`, which you can then split into `(x + 1)/(x + 1) - 1/(x + 1)`, which you can then integrate much more easily.
- markisus 3y ago
  Why not directly u substitute u = x+1
- Spivak 3y ago
  I liked these problems because they make you feel clever figuring them out but hated them because they were a mountain of tedious algebra.
mydogcanpurr 3y ago
The approach is less mysterious if you recognize it as going back to definitions, which is a common problem solving technique.
ogogmad 3y ago
One technique for finding the derivative of the sin function is to find the derivative of arcsin first. The arcsin function can be expressed as the area of a certain figure, and therefore admits an expression as an integral. From this, the derivative of arcsin is immediate. Finally, apply your technique to arcsin(sin(x)) = x and obtain the derivative of sin. A similar technique finds the derivative of exp(x) from ln(x), by defining the latter as the integral of 1/x.
- selimthegrim 3y ago
  ChatGPT?
  - ogogmad 3y ago
    :P See https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/1230225#1230225 https://math.stackexchange.com/questions/75130/how-to-prove-... The OP uses the derivative of sin to find the derivative of arcsin. My point was that the other way round is sort of natural as well. Thought I'd share. The broader point is that you can obtain all the classic definitions and identities of the elementary functions [1] by combining only (i) the four arithmetic operations, (ii) integration, and (iii) taking inverses of functions (iv) the constants 0 and 1. I guess that's a bit cool. This works fine over the real numbers. Over the complex numbers, this might require Riemann surfaces to work. e^z in complex analysis is usually defined by its Taylor series, instead of as the inverse function of the natural log (whose domain is actually a Riemann surface). [1] - https://en.wikipedia.org/wiki/Elementary_function https://en.wikipedia.org/wiki/Elementary_function
mjd 3y ago
This is a great example. Thanks very much!
justeleblanc 3y ago
That's how you find any reciprocal function's derivative. If g = f^{-1}, then g'(y) = 1/f'(g(y)).
- Spivak 3y ago
  By your rule g = arcsin f = sin g'(x) = 1 / cos(arcsin(x)) But that's not the usual form. Can you show that the denominator is sqrt(1 - x^2)? Doing that requires you make the same observation as in the parent post.
  - Tainnor 3y ago
    > Can you show that the denominator is sqrt(1 - x^2)? That's a trig identity that is easily derivable: draw a right triangle with hypotenuse 1 and one side being x. Its one angle will be arcsin(x) by definition. Then by Pythagoras the remaining side is sqrt(1-x^2), which is the cosine of that angle. I guess GP's point was that the implicit differentiation trick by using the inverse is generalisable and not only restricted to trigonometric functions. Obviously, which simplifications you can then additionally make will depend on the function in question.

tonyarkles 3y ago

Alternatively, x = sin(theta) and arcsin(sin(theta)) = theta, depending on the limits. It’s a cool technique, thanks for the reminder!

77pt77 3y ago

Dude. That's just a direct application of the inverse function theorem. Even geometrically you can see that swapping the axis (x and y) gives you the desired result.

CGamesPlay 3y ago
GP just happened to be one of the lucky 10,000 on that day in math class. https://xkcd.com/1053/ https://xkcd.com/1053/

saghm 3y ago

In turn, this reminds me a bit of a calculus problem I saw freshman year of college that I ended up sharing with my old high school calculus teacher because it similarly looks intractable until you make a "simplification". The problem was to find the integral of `x/(x + 1)` (I forget the exact bounds, presumably from 0 to x with respect to x). The trick is that this the same as `(x + 1 - 1)/(x + 1)`, which you can then split into `(x + 1)/(x + 1) - 1/(x + 1)`, which you can then integrate much more easily.

markisus 3y ago
Why not directly u substitute u = x+1
Spivak 3y ago
I liked these problems because they make you feel clever figuring them out but hated them because they were a mountain of tedious algebra.

mydogcanpurr 3y ago

The approach is less mysterious if you recognize it as going back to definitions, which is a common problem solving technique.

ogogmad 3y ago

One technique for finding the derivative of the sin function is to find the derivative of arcsin first. The arcsin function can be expressed as the area of a certain figure, and therefore admits an expression as an integral. From this, the derivative of arcsin is immediate. Finally, apply your technique to arcsin(sin(x)) = x and obtain the derivative of sin. A similar technique finds the derivative of exp(x) from ln(x), by defining the latter as the integral of 1/x.

selimthegrim 3y ago
ChatGPT?
- ogogmad 3y ago
  :P See https://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1/1230225#1230225 https://math.stackexchange.com/questions/75130/how-to-prove-... The OP uses the derivative of sin to find the derivative of arcsin. My point was that the other way round is sort of natural as well. Thought I'd share. The broader point is that you can obtain all the classic definitions and identities of the elementary functions [1] by combining only (i) the four arithmetic operations, (ii) integration, and (iii) taking inverses of functions (iv) the constants 0 and 1. I guess that's a bit cool. This works fine over the real numbers. Over the complex numbers, this might require Riemann surfaces to work. e^z in complex analysis is usually defined by its Taylor series, instead of as the inverse function of the natural log (whose domain is actually a Riemann surface). [1] - https://en.wikipedia.org/wiki/Elementary_function https://en.wikipedia.org/wiki/Elementary_function

mjd 3y ago

This is a great example. Thanks very much!

justeleblanc 3y ago

That's how you find any reciprocal function's derivative. If g = f^{-1}, then g'(y) = 1/f'(g(y)).

Spivak 3y ago
By your rule g = arcsin f = sin g'(x) = 1 / cos(arcsin(x)) But that's not the usual form. Can you show that the denominator is sqrt(1 - x^2)? Doing that requires you make the same observation as in the parent post.
- Tainnor 3y ago
  > Can you show that the denominator is sqrt(1 - x^2)? That's a trig identity that is easily derivable: draw a right triangle with hypotenuse 1 and one side being x. Its one angle will be arcsin(x) by definition. Then by Pythagoras the remaining side is sqrt(1-x^2), which is the cosine of that angle. I guess GP's point was that the implicit differentiation trick by using the inverse is generalisable and not only restricted to trigonometric functions. Obviously, which simplifications you can then additionally make will depend on the function in question.