# Question #a254f

Feb 16, 2018

It should be $\frac{d}{\mathrm{dx}} \left({\sin}^{-} 1 x\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

#### Explanation:

$y = {\sin}^{-} 1 x$ so $\sin y = x$.

Now use implicit differentiation to get

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} y$

and use trigonmetry to get

$= \frac{1}{\sqrt{1 - {x}^{2}}}$

Feb 17, 2018

A more visual yet same as the answer below

#### Explanation:

$y = {\sin}^{-} 1 \left(x\right) \iff \sin \left(y\right) = x$

$\sin \left(y\right) = x$

Create a triangle where $\sin \left(y\right) = \frac{x}{1}$ The missing side can be found using Pythagorean's Theorem

Differentiate both sides implicitly (W.R.T. $x$)

$\frac{\mathrm{dy}}{\mathrm{dx}} \cdot \cos \left(y\right) = 1$

Divide $\cos \left(y\right)$ from both sides

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right)$

Rewrite in terms of $x$

Since $\sin \left(y\right) = \frac{x}{1}$

Then $\cos \left(y\right) = \frac{\sqrt{1 - {x}^{2}}}{1} = \sqrt{1 - {x}^{2}} \leftarrow$ Based on triangle

So $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{\cos} \left(y\right) = \frac{1}{\sqrt{1 - {x}^{2}}}$

$\therefore \frac{\mathrm{dy}}{\mathrm{dx}} \left[\arcsin \left(x\right)\right] = \frac{1}{\sqrt{1 - {x}^{2}}}$