🗑️ 🧵 Why does this method fail?
Anonymous at Fri, 19 Jul 2024 17:45:12 UTC No. 16288607
So in calculus class we were looking at the proofs of the derivatives of the inverse trigonometric functions. It went something like this:
Suppose we have [math]y = \arcsin x[/math], then we can take [math]\sin[\math] on both sides and we get [math]\sin y = x[/math]. Then we can use implicit differentiation to get: [math]\cos y \frac{dy}{dx} = 1[/math], then [math]\frac{dy}{dx} = \frac{1}{\cos y}[/math]. From here we can use the fact that [math]\sin y = x[/math] and imagine a right triangle where one of the acute angles is [math]y[/math], the side opposite to that angle is [math]x[/math] and the hypotenuse is [math]1[/math]. This means that the missing side is equal to [math]\sqrt{1-x^2}[/math], and therefore [math]\cos y = \frac{\sqrt{1-x^2}}{1}[/math]. If we substitute that we get [math]\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}[/math], which is exactly what you should get. You can continue like this and you would get all the other ones right, except for [math]\sec^{-1}[/math] and [math]\csc^{-1}[/math]. If you follow the method, you get: [math]\frac{d}{dx}\sec^{-1}x = \frac{1}{x \sqrt{x^2-1}}[/math] and [math]\frac{d}{dx}\cos^{-1}x = \frac{-1}{x \sqrt{x^2-1}}[/math], but that is not the correct result, as you can see is missing the absolute value. So what gives? Why does the method fail for some but it's correct for others?
Anonymous at Fri, 19 Jul 2024 17:45:42 UTC No. 16288609
>we
Anonymous at Fri, 19 Jul 2024 18:02:26 UTC No. 16288628
>homework thread
Anonymous at Fri, 19 Jul 2024 18:05:25 UTC No. 16288632
>>16288609
Yes, we. You were there as well
>>16288628
What? It was a calculus class, do you really think they'd assign "find the error in this proof" as homework? It's just something that was never resolved in class, I asked the teacher and he didn't know how to answer.
Anonymous at Fri, 19 Jul 2024 18:18:45 UTC No. 16288648
Your question is poorly phrased. Rewrite it to be more succinct and I might read it.
Anonymous at Fri, 19 Jul 2024 19:10:55 UTC No. 16288702
>>16288607
I am too lazy atm to figure out exactly why, but I assume it has to do with the domain of functions like arcsine being [-1,1] while for arcsecant it is (-inf, -1] U [1, inf). The latter function has two separate branches, so you have to consider the derivative on each branch. What you think should be the answer is probably secretly only considering one branch, like [1, inf).
Anonymous at Fri, 19 Jul 2024 19:30:23 UTC No. 16288728
>>16288607
>as you can see is missing the absolute value
the hypotenuse is always positive
Anonymous at Fri, 19 Jul 2024 19:47:59 UTC No. 16288755
>>16288728
>the hypotenuse is always positive
Even with complexe numbers?
Anonymous at Fri, 19 Jul 2024 21:06:50 UTC No. 16288858
>>16288755
Complex numbers is something else. The Pythagorean theorem assumes real values. If you apply complex number, you shouldn't expect the theorem to be true.
Anonymous at Fri, 19 Jul 2024 21:11:17 UTC No. 16288863
>>16288755
Yes, but at that point it's a 2-norm not the Pythagorean theorem (though it's basically the same thing).
Anonymous at Fri, 19 Jul 2024 21:27:37 UTC No. 16288884
>>16288632
>Teacher couldn't figure it out
Damn the bar is LOW. Tell me more about your demographics. State, income ranges in your area, teacher's degree, public/private school, ethnic and religious makeup of your area, red/blue/swing/third party, etc. I want to know ALL that heckin /pol/ data.
>>16288702
Has the right idea.
Anonymous at Fri, 19 Jul 2024 21:57:28 UTC No. 16288922
>>16288607
I'll do it for arcsec. We have u = sec y and by the method you understand
[math] \frac{dy}{du} = \frac{\cos^2 y}{\sin y} = \frac{1}{u^2 \sin y} [/math]
There's no ambiguity up to this point, but now we have to know if \sin y is positive or negative. The standard definition of arcsec is such that y ranges from from 0 to \pi, so \sin y is positive:
[math] \frac{dy}{du} = \frac{1}{u^2\sqrt{1-u^{-2}} =\frac{1}{|u|\sqrt{u^2-1} [/math]
Since we can only factor positive numbers out of the square root.
Anonymous at Fri, 19 Jul 2024 21:59:02 UTC No. 16288924
Oops here's the rendered math above
[math]\frac{dy}{du} = \frac{1}{u^2\sqrt{1-u^{-2}}} =\frac{1}{|u|\sqrt{u^2-1}} [/math]
Anonymous at Sat, 20 Jul 2024 00:42:25 UTC No. 16289065
>>16288607
It's only true on a certain domain. For the arc sec and csc, it's always either + or - the value. Dude, just graph arc sec and csc. Is the slope of arcsec always positive? Is it always negative for arccsc? No, of course not. Look up what domain is commonly used for these derivatives.
As for WHY they're + or -, do the derivation yourself, and notice you're doing a sinx^2 = 1-cos^2x or similar. That means sinx is either plus or minus the square root of the LHS. You know if it's plus or minus if x is within one domain or another domain, like between 0 and pi