# What is the derivative of \sin^{-1}(x)

What is the derivative of ${\mathrm{sin}}^{-1}\left(x\right)$
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abonirali59
Let $y={\mathrm{sin}}^{-1}\left(x\right)$
Thus, $\mathrm{sin}\left(y\right)=x$ and $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$
Now, diferentiate implicity
$\mathrm{cos}\left(y\right)\frac{dy}{dx}=1$, so
$\frac{dy}{dx}=\frac{1}{\mathrm{cos}\left(y\right)}$
As we know, that $-\frac{\pi }{2}\le y\le \frac{\pi }{2}$, we understand that $\mathrm{cos}\left(y\right)$ is positive
$\frac{dy}{dx}=\frac{1}{\sqrt{1-{\mathrm{sin}}^{2}\left(y\right)}}=\frac{1}{\sqrt{1-{x}^{2}}}$
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Ethan Sanders
${\mathrm{sin}}^{-1}\left(x\right)=\mathrm{arcsin}\left(x\right)$
$\frac{d}{dx}\left(\mathrm{arcsin}\left(x\right)\right)$
Apply the common derivative
$=\frac{1}{\sqrt{1-{x}^{2}}}$