Comment on Activity 7

The third side of the triangle in Figure 4 is ${\displaystyle \sqrt{1-x^2}}$, so this is also ${\displaystyle cos y}$.

${\displaystyle \frac{d}{dx}sin y}$	${\displaystyle =\frac{d}{dx} x}$
${\displaystyle cos y \frac{dy}{dx}}$	${\displaystyle =1}$
${\displaystyle \frac{dy}{dx}}$	${\displaystyle =\frac{1}{cos y}}$
${\displaystyle \frac{dy}{dx}}$	${\displaystyle =\frac{1}{\sqrt{1-x^2}}}$

We differentiate implicitly the formula ${\displaystyle sin y=x}$, where ${\displaystyle y={sin}^{-1}x:}$${\displaystyle }$

${\displaystyle \frac{dy}{dx}}$	${\displaystyle =\frac{1}{\frac{dx}{dy}}}$
	${\displaystyle =\frac{1}{\frac{d}{dy} sin y}}$
	${\displaystyle =\frac{1}{cos y}}$
	${\displaystyle =\frac{1}{\sqrt{1-x^2}}}$

Thus, ${\displaystyle \frac{d}{dx} {sin}^{-1}x=\frac{1}{\sqrt{1-x^2}}}$. Now we see that the inverse function formula leads to the same result: