Blaze Jensen

2023-03-14

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

dhopeefreshnwn

Beginner2023-03-15Added 7 answers

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.

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.