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25 Dec 2021Mathchan is launched into public
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Image
  No. 1099
>>1100
It is both kinda surprising and disappointing that the easiest way to compute the derivative of
arc sinarc\ sin
is to solve the integral of
1−x2\sqrt{1-x^2}
geometrically.

Just by looking on a quarter of a circle with angle in it you know that:
∫1−x2dx=arc sinx2+x1−x22\int{\sqrt{1-x^2}}dx = \frac{arc\ sinx}{2} + \frac{x\sqrt{1-x^2}}{2}


By deriving everything we get:
1−x2=(arc sinx)′2+(x1−x2)′2\sqrt{1-x^2} = \frac{(arc\ sinx)'}{2} + \frac{(x\sqrt{1-x^2})'}{2}


And now it's easy to get
(arc sinx)′(arc\ sinx)'
out of the equation and solve it.

It is
11−x2\frac{1}{\sqrt{1-x^2}}
if anyone asked.

Then you can compute
(arc cosx)′(arc\ cosx)'
:

(arc cosx)′=(arc sin(1−x))′=(arc\ cosx)' = (arc\ sin(1-x))' =

=(1−x)′11−(1−x)2=...= (1-x)'\frac{1}{\sqrt{1-(1-x)^2}} = ...


Similarily and even easier
(arc tanx)′(arc\ tanx)'
because it's just a derivative of a fraction of functions.
>>
  No. 1100 >>1102

File: Ears.gif ( 357.71 KB , 425x599 , 1740956279555.gif )

Image
>>1099

the inverse function of any differentiable bijection can have
its derivative computed via the chain rule. just differentiate their composition.

gross wojack picture btw
>>
  No. 1102
>>1100
Yes, now I see. I suspected that there should be some easier way to do this because these are inverse functions.

This wojack is a homage to some Serbian who was shilling Lukyon on the soy wojak forum. He also used this "neutral feraljak".

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