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