 Blaze Jensen

2023-03-14

What is the derivative of $y=\mathrm{arcsin}\left(x\right)$? dhopeefreshnwn

This identity can be proven easily by applying sin to both sides of the original equation:
1.) $y=\mathrm{arcsin}x$
2.) $\mathrm{sin}y=\mathrm{sin}\left(\mathrm{arcsin}x\right)$
3.) $\mathrm{sin}y=x$
We continue by using implicit differentiation, keeping in mind to use the chain rule on sin y:
4.) $\mathrm{cos}y\frac{dy}{dx}=1$
Solve for $\frac{dy}{dx}$:
5.) $\frac{dy}{dx}=\frac{1}{\mathrm{cos}y}$
Then, substitution with our original equation yields $\frac{dy}{dx}$ in terms of x:
6.) $\frac{dy}{dx}=\frac{1}{\mathrm{cos}\left(\mathrm{arcsin}x\right)}$
This may not appear to be a good idea at first, but it can be simplified if the identity is remembered.
$\mathrm{sin}\left(\mathrm{arccos}x\right)=\mathrm{cos}\left(\mathrm{arcsin}x\right)=\sqrt{1-{x}^{2}}$.
7.) $\frac{dy}{dx}=\frac{1}{\sqrt{1-{x}^{2}}}$
This is a good definition to memorize, along with $\frac{d}{dx}\left[\mathrm{arccos}x\right]$ and $\frac{d}{dx}\left[\mathrm{arctan}x\right]$, since they appear quite frequently in differentiation problems.

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