Derive $\frac{d}{dx} \left[\sin^{-1} x\right] = \frac{1}{\sqrt{1-x^2}}$

Solution 1:

Edited in response to Aryabhata's comment. From $\cos ^{2}y+\sin ^{2}y=1$, we get $\cos y=\pm \sqrt{1-\sin ^{2}y}$. For $y\in \lbrack -\pi /2,\pi /2]$, $\cos y=\sqrt{1-\sin ^{2}y}\geq 0$, and $% y=\arcsin x\Leftrightarrow x=\sin y$ (see Inverse trigonometric functions). Then by the rule of the inverse function we have

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

Solution 2:

You are on the right track. All that's left is to write $\sec(y)$ in terms of $x$. To do this, recall that $y=\arcsin(x)$. That is, $y$ is some angle, sine of which yields $x$. Thus, there is a right triangle with angle $y$, the side opposite $y$ has length $x$ and the hypotenuse is $1$. You need to compute the secant of this angle $y$.

arcsin http://homepage.mac.com/shelleywalsh/MathArt/arcsin.gif

This approach generalizes to finding the derivatives of the other inverse trig functions, and is a good way to wrap your head around composing trig functions and inverse trig functions of any flavor.