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Derivative of arcsin(x)

Petrus

Well-known member
Feb 21, 2013
739
Hello MHB,
derivate \(\displaystyle \sin^{-1}(x)\)


So I use the derivate formula for invers and get
\(\displaystyle \frac{1}{\cos(\sin^{-1}(x))}\)
and Then draw it and get \(\displaystyle \frac{1}{\sqrt{1-x^2}}\)
but there is a reason WHY it cant be \(\displaystyle -\frac{1}{\sqrt{1-x^2}}\) and I did not understand it, I did not get it.

Regards,
\(\displaystyle |\pi\rangle\)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,888
Re: Derivate of arcsin(x)

Hello MHB,
derivate \(\displaystyle \sin^{-1}(x)\)


So I use the derivate formula for invers and get
\(\displaystyle \frac{1}{\cos(\sin^{-1}(x))}\)
and Then draw it and get \(\displaystyle \frac{1}{\sqrt{1-x^2}}\)
but there is a reason WHY it cant be \(\displaystyle -\frac{1}{\sqrt{1-x^2}}\) and I did not understand it, I did not get it.

Regards,
\(\displaystyle |\pi\rangle\)
The $\arcsin$ is defined to have a range of $-\pi/2$ to $+\pi/2$.
With this definition the derivative is always positive.
You can also choose the range to be different, making the derivative negative, but then it's not an $\arcsin$ anymore. Then you have a different inverse for the sine.