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# 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.

{$\color{crimson}{y = \arcsin(x)}$}, {$\color{forestgreen}{y=\sin(x)}$}, and {$\color{blue}{y = d/dx \arcsin(x)}$}

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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Categories: DerivativeDemonstrations

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