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Derive $\frac{d}{dx} \left[\sin^{-1} x\right] = \frac{1}{\sqrt{1-x^2}}$ (Hint: set $x = \sin y$ and use implicit differentiation)

So, I tried to use the hint and I got:

$x = \sin y$

- Closed and exact.
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$\frac{d}{dx}\left[x\right] = \sin y\frac{d}{dx}$

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

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

$\frac{dy}{dx} = \sec y$

From here I need a little help.

- Did I do the implicit

differentiation correctly? - How do I use this to help with

the original question?

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*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

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

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.

You just need the fact that $\cos(x) = \sqrt{1-\sin^2(x)}$ and you should be able to complete it from what you already have.

You are nearly there. You just need to take $\frac{dy}{dx} = \frac{1}{\cos y}$ and express both sides in terms of $sin(x)$ (LHS) and $x$ RHS

LHS will give you LHS of your equation. And remembering $x=sin(y)$ and a formula for $cos(y)$ in terms of $sin(y)$ will get you the RHS. $sec(y)$ is a distraction best avoided.

First, note that $\displaystyle \sin^{-1}: [-1,1] \to [-\frac{\pi}{2}, \frac{\pi}{2}]$

The range is important, as for this range, you have that if $y = \sin^{-1} x$ then $\cos y = \sqrt{1 – x^2}$, as we have $\sin y = x$ *and* $\cos y \ge 0$ whenever $y \in [-\frac{\pi}{2}, \frac{\pi}{2}]$.

i.e. since $\sin^2 y + \cos^2 y = 1$, we get $\cos y = \pm \sqrt{1 – x^2}$ and since $\cos y \ge 0$, we can say $\cos y = \sqrt{1- x^2}$.

**Exercise**

Suppose we defined $\sin^{-1}x$ as the unique angle $\theta$ in $[\frac{\pi}{2}, \frac{3 \pi}{2}]$ such that $\sin \theta = x$, what is the derivative of $\sin^{-1} x$?

You can also use implicit diff. Begin with

$$y = \sin^{-1}(x);$$

Apply the sine function to get

$$\sin(y) = x.$$

Now differentiate.

$$\cos(y) y' = 1,$$

so

$$y' = {1\over \cos(y)}.$$

Now draw the $1$-$x$-$\sqrt{1 – x^2}$ triangle and compute and you get

$$y' = {1\over\sqrt{1 – x^2}}.$$

This technique works nicely for various inverse functions.

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