y=arcsin(x), what is y' ?
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,216 Nov 20, 2011 #2 Start with \(\displaystyle x=sin(y)\Rightarrow y=sin^{-1}(x)\) Differentiate: \(\displaystyle 1=cos(y)\frac{dy}{dx}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{cos(y)}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{cos(sin^{-1}(x))}\) But, \(\displaystyle cos(sin^{-1}(x))=\sqrt{1-x^{2}}\) \(\displaystyle \therefore \;\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}\)
Start with \(\displaystyle x=sin(y)\Rightarrow y=sin^{-1}(x)\) Differentiate: \(\displaystyle 1=cos(y)\frac{dy}{dx}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{cos(y)}\) \(\displaystyle \frac{dy}{dx}=\frac{1}{cos(sin^{-1}(x))}\) But, \(\displaystyle cos(sin^{-1}(x))=\sqrt{1-x^{2}}\) \(\displaystyle \therefore \;\ \frac{dy}{dx}=\frac{1}{\sqrt{1-x^{2}}}\)
