# Derivative of Arcsin Demonstration

**Arcsin**

Demonstration of the derivative of Arcsin

Arcsin is defined: {$$ \large \arcsin(x) = y \iff \sin(y) = x $$}

with the usual range being:

{$$ \large -{\pi \over 2}\le y \le {\pi \over 2} $$}

Or in other words, we are dealing with the first and third quadrants here. An implication of defining the range this way is that {$\cos(y)$} is always positive.

This demonstration proceeds by implicit differentiation.

{$$ \large \begin{align} \arcsin(x) = y &\iff \sin(y) = x \tag{definition} \cr \sin(y) &= x \cr \left( \sin(y) \right) ^\prime &= \left( x \right) ^\prime \cr {d \over {dx}}\sin(y)y^\prime &= 1 \tag{Chain rule} \cr \cos(y)y^\prime &= 1 \cr y^\prime &= {1 \over {\cos(y)}} \cr y^\prime &= {1 \over {\sqrt{1 - \sin^2(y)}} } \tag{Pythagorean}\end{align} $$}

Taking only the positive root because {$\cos(y)$} is nonnegative in the first and third quadrants. Substituting back for {$x$}:

{$$ \Large \therefore \arcsin^\prime x = {1 \over {\sqrt{1 - x^2}} } $$}

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