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

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:

$x = \sin y$

$\frac{d}{dx}\left[x\right] = \sin y\frac{d}{dx}$

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

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

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

From here I need a little help.

1. Did I do the implicit differentiation correctly?
2. How do I use this to help with the original question?
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The notation in the second step is not right. You need to write $\frac{d}{dx}\sin y$, not $\sin y \frac{d}{dx}$. (The derivative operator acts on what comes after it). –  Hans Lundmark Jun 30 '11 at 22:38
It's pretty well covered in existing answers, but my answer on this other question shows a technique for organizing the algebra/trigonometry involved in getting from $x=\sin y$ and $\sec y$ to $\frac{1}{\sqrt{1-x^2}}$, though demonstrated with a different pair of functions. –  Isaac Jun 30 '11 at 22:39

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

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

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Ok, I can understand where this is going, I just didn't know you could inverse $\displaystyle \frac{dy}{dx}$ like that... –  OghmaOsiris Jun 30 '11 at 20:43
@OghmaOsiris: I added a link to the Wikipedia entry. –  Américo Tavares Jun 30 '11 at 20:46
Nitpick: $y = \arcsin x \Leftrightarrow x = \sin y$ is incorrect. You need to mention that $y \in [-\pi/2, \pi/2]$. You also need to use that fact to be able to state $\cos y = \sqrt{1 - \sin^2 y}$ (i.e. eliminate the possibility $\cos y = - \sqrt{1 - \sin^2 y}$). –  Aryabhata Jun 30 '11 at 22:15
@Aryabhata: I knew that, but I didn't explained it properly. Thanks! I edited my answer. –  Américo Tavares Jun 30 '11 at 22:26

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 $y$.

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.

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@wckronholm if would be better if could draw the pic ;) –  timhortons Jun 30 '11 at 20:15
@IDIOT I agree. I'll figure out how to include images in my answers and edit to include the appropriate triangle. –  wckronholm Jun 30 '11 at 20:16
@wckronholm very nice! –  timhortons Jun 30 '11 at 20:24
@OghmaOsiris If you think of the triangle trig definitions of the trig functions, then secant is hypotenuse/adjacent. Does that help? –  wckronholm Jun 30 '11 at 20:51
@OghmaOsiris I, of course, also don't know if this is the proof your teacher wants, or even if there is one proof in particular your teacher expects. The technique I described above is (I think) fairly standard, and it's something I show my students when I teach differential calculus. –  wckronholm Jun 30 '11 at 20:58

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.

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$\cos x = \sqrt{1 - \sin^2 x}$ is not always right... –  Aryabhata Jun 30 '11 at 22:19
@Aryabhata: Ah, yes, you are correct. Bill has already corrected his answer so I'm just gonna leave mine the way it is. –  El'endia Starman Jul 1 '11 at 1:51

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.

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Ok... what? does LHS = left hand side? –  OghmaOsiris Jun 30 '11 at 20:29
LHS - left hand side, RHS = right hand side. –  Mark Bennet Jun 30 '11 at 20:32

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 $\sin^{-1} x$?

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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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