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### Course: Class 12 math (India) > Unit 6

Lesson 3: Inverse trig functions differentiation# Derivative of inverse sine

Unlock the mystery of the derivative of inverse sine! Let's dive into the world of calculus, rearranging equations and applying implicit differentiation to find the derivative of y with respect to x. Using trigonometric identities, we transform the derivative into a function of x, revealing a fascinating relationship. Created by Sal Khan.

## Want to join the conversation?

- At3:05: What assures us that it shouldn't be plus and minus the square root? Like this: cos(y)=+-sqrt(1-sin(y)^2)?(184 votes)
- Good question! Sal skipped over the reason, and it's far from obvious.

The answer has to do with the way we define the inverse sine function. As you know, the sine function is cyclical, so we wouldn't get a valid function if we created its inverse in the usual way of simply reflecting it across the y = x diagonal because we wouldn't have a unique output for valid inputs. For example, the inverse sine of 0 could be 0, or π, or 2π, or any other integer multiplied by π. To solve this problem, we restrict the range of the inverse sine function, from -π/2 to π/2. Within this range, the slope of the tangent is always positive (except at the endpoints, where it is undefined). Therefore, the derivative of the inverse sine function can't be negative. We can safely discard the negative square root in this derivation because it would give us a negative derivative, which is impossible because of the way we've restricted the domain of the inverse sine function.(281 votes)

- I am a bit rusty with taking derivatives of inverse functions but is there a reason why when taking the derivative of y=(sin x)^-1 that we could not use the Power or Quotient Rule to solve the function? When I use the power/quotient rule I get d(y)/dx = d((sin x)^-1)/dx = -cos x/(sin x)^2 and not 1/sqrt(1-x^2) that Sal mentions. Where am I going wrong?(52 votes)
- The inverse functions, though written as sin⁻¹, etc. ARE NOT the reciprocals of those functions. They are NOT being raised to the -1 power. Thus, what you were doing was finding the derivatives of the reciprocal functions, not the inverse functions.

So, remember that sin⁻¹ x is NOT (sin x)⁻¹ and is NOT 1 / sin x.

To avoid confusion, you can use the alternative notation of arc-

sin⁻¹ x = arcsin x

The same goes for all of the other trig functions.(169 votes)

- Hi everyone, maybe I'am asking this question because I just had not watched the videos about trigonometry but maybe someone can give me a quick explanation. So, why y=sin^(-1)x is the same as siny=x(49 votes)
- Suppose
*y=x²*. This is y in terms of x.

Now if you want to find out what x is in terms of y, then solve for x to get*x=√y*.

As you know, the square operator and the square root operator are inverses of each other, that is, one "undoes" the other:*√(x²) = (√x)² = x*(assuming we are only interested in the*principal*square root).

It is the same deal with*sin*and*arcsin*, which is conventionally written as*sin^-1 x*.*Arcsin*is the inverse of*sin*, such that*arcsin(sin(x)) = x*, or*sin(arcsin(x))=x*.

Like the square/square root example, if you have*y=sin(x)*, which is y in terms of x, but you want to take that expression and find x in terms of y, then given:*y=sin(x)*

take the arcsin of both sides:*sin^-1(y)=sin^-1(sin(x))*, so that:*sin^-1(y)=x*

or, using the term*arcsin*and not*sin^-1*(though*sin^-1*is more common)*y=sin(x)**arcsin(y) = arcsin(sin(x)) = x*, so that*arcsin(y) = x*

It is important to know the inverse trig functions as they come in handy in many situations, like trig substitution in integral calculus.(80 votes)

- At2:40, you begin to explain that you want to change dy / dx = 1 / cos y into something in terms of x by using the Pythagorean Identity and setting that (the Pythagorean Identity) equal to cos y . Could you just have inserted your original definition of y (y = sin^-1 x) instead? You would end up with:

dy / dx = 1 / (cos (sin^-1 x)).

Does that work ? tried it with a few arbitrary numbers for x and was getting the same answer for both equations, but wasn't sure if it technically was correct.(20 votes)- Yes. If you want to visualize, try to type
`cos(arcsin(x))`

and`sqrt(1-x^2)`

into Google or`is cos(arcsin(x)) equal to sqrt(1-x^2)`

into WolframAlpha.(13 votes)

- When prompted in the beginning, I solved dy/dx for (sinx)^-1, and got a different result.

I looked into it and realized that (sinx)^-1 and sin^-1(x) are different things.

How come is (sinx)^-1 =/= sin^-1(x) but (sinx)^2 = sin^2(x) ?

Did someone deliberately come up with notations that are as confusing as possible just to mess with me or am I missing something?(12 votes)- Inverse functions are, perhaps unfortunately, often denoted with the superscript
`-1`

. This can be confusing at times if it is not made clear whether we mean to denote the inverse function or the reciprocal function. However, the inverse trigonometric functions, known as the*arcus*functions, are typically denoted`arcsin`

,`arccos`

, etc. So instead of writing the inverse of the sine function at`x`

as`sin^(-1) x`

, one can also write`arcsin x`

, which is better practice, I would argue.(17 votes)

- I'm a little rusty on inverse trig functions, but why is it that he can write y=sin^(-1)x as sin(y)=x?(7 votes)
- First and foremost, know that the inverse sine function is not the sine function to the negative 1 power. It's better if you write out inverse sine rather than sine to the negative one power. But to answer your question, if y=arcsin(x) which means the same thing as inverse sine, then sin(y)=x. That was the first step of the problem.(11 votes)

- Is there any video on Khan Academy about the derivatives and integrals of hyperbolic functions? Another question: Can complex numbers be differentiated or integrated?(4 votes)
- The hyperbolic functions are usually defined in terms of the exponential function. As such, their properties are easily deduced from properties of the exponential function.

Complex-valued functions defined on a subset of complex numbers may (some times) be differentiated, yes. Let

denote the set of complex numbers, and suppose**C**`U`

is some subset of

. Suppose**C**`ƒ: U →`

is a complex-valued function defined on**C**`U`

, and suppose`w`

is an interior point of`U`

. If the limit`lim (z → w) [ƒ(z) - ƒ(w)] / [z - w]`

exists and is finite, we say that`ƒ`

is differentiable at`w`

, and we denote the value of this limit by`ƒ'(w)`

. If`U`

is open, and if`ƒ`

is differentiable on all of`U`

, we say that`ƒ`

is differentiable (on`U`

). If`ƒ`

is differentiable on an open set`U`

, one also says that`ƒ`

is*holomorphic*on`U`

, or sometimes that`ƒ`

is*analytic*on`U`

. Holomorphic functions are central in the theory of complex functions.

More specifically, to say that`ƒ: U →`

is differentiable at an interior point**C**`w`

in`U`

means the following: there exists some complex number`L`

such that for every real number`ε > 0`

there exists a real number`δ > 0`

with the property that for all complex numbers`z`

in`U`

with`0 < |z - w| < δ`

, we have`|[ƒ(z) - ƒ(w)]/[z - w] - L| < ε`

. If such a number`L`

exists, we usually denote it by`ƒ'(w)`

. This property may also be cast in terms of convergent sequences in`U`

.

The process of differentiation of complex-valued functions defined on subsets of the complex numbers share many properties with differentiation of real-valued functions defined on subsets of the real numbers. For instance, the differentiation operator is linear. Furthermore, the product rule, the quotient rule, and the chain rule all hold for such complex functions.

I will not include a discussion on integration of complex-valued functions defined on subsets of

, as this would require more sophisticated typesetting than what is available here.**C**(4 votes)

- Do mathematician needs cheat sheet while solving calculus?(5 votes)
- Derivatives? No, everyone generally does it from memory. Integral tables? Yes. No one has the time to memorize thousands of different forms, even if they are less commonly used.(7 votes)

- Once one gets to dy/dx = cos^-1(y), could the original identity y=sin^-1(x) be substituted back in? So it would be dy/dx = cos^-1(sin^-1(x))? Is this equivalent? Thanks.(3 votes)
- Your idea is correct, but your final answer is not.

Just to remember:`cos⁻¹(x)`

is NOT equal to`(cos(x))⁻¹`

.`cos⁻¹(x)`

is equal to`arccos(x)`

The correct answer would be:`dy/dx = cos(sin⁻¹(x))⁻¹`

(5 votes)

- Is there anything like an inverse derivative? For example, the inverse derivative of cos(x) would be sin(x)? Thank you. Please also provide a detailed explanation. Again, thanks in advance!(3 votes)
- Just like addition and subtraction are inverse functions, multiplication and division are inverse functions, and derivatives and integrals are inverse functions. Take a look at integral calculus to learn more about integrals. :) It's not quite as cut-and-dry as the other inverse operations (think about the derivative of f(x)=x^2+5 being f'(x)=2x, and you realize that going backward you wouldn't know there would be a '+5').(3 votes)

## Video transcript

What I would like to explore in this
video, is to see if we could figure out the derivative of Y is
with respect to X. If Y is equal to the inverse sine, the
inverse sine of X. And like always, I encourage you to pause
this video and try to figure this out on your
own. And I will give you two hints. First hint is, well, we don't know what
the derivative of sine inverse of x is, but we do know what the si-, what the
derivative of the sine of something is. And so if you, maybe if you rearrange this
and use some implicit differentiation, maybe you
can figure out what dy, dx is. Remember, this is right over here. This right over here is our goal. But, since you want to figure our the
derivative of this with respect to x. So, assuming you've had a go at it, so
let's work through this together. So, if y is the inverse sine of x, that's
just like saying that, that's equivalent to
saying that sine of y. Sine of Y is equal to X. Sine of Y is equal to X. So now we have things that we're a little
bit more familiar with, and now we can do a little bit of
implicit differentiation. We can take the derivative of both sides
with respect to X. So, derivative of the left-hand side with
respect to X and the derivative of the right-hand side
with respect to X. But what's the derivative of the left-hand
side with respect to X going to be? And here we just apply the chain rule. It's going to be the derivative of sine of
Y with respect to Y. Which is going to be, which is going to be
cosine of Y times the derivative of Y with
respect to X. So times dy, dx, times dy,dx. And the right-hand side, what's the
derivate of X with respect to X, well that's obviously just going to
be equal to one. And so we could solve for dy,dx, divide
both sides by cosine of Y. And we get the derivative of Y with
respect to X is equal to one over cosine of Y. Now this still isn't that satisfying cuz I
have the derivative in terms of Y. So let's see if we can re-express it in
terms of X. So, how could we do that? Well, we already know that X is equal to
sine of Y. Let me rewrite it. We already know that X is equal to sine of
Y. So, if we could rewrite this bottom expression in terms, instead of cosine of
Y. If we could use our trigonometric
identities to rewrite it in terms of sine of Y, then we'll be in good shape because X
is equal to sine of Y. Well, how can we do that? Well, we know from our trigonometric
identities, we know that sine squared of Y plus cosine squared
of Y is equal to one. Or, if we want to solve for cosine of Y, subtract sine squared of Y from both
sides. We know that cosine squared of Y is equal
to one minus sine squared of Y. Or that cosine of Y, just take the
principal root of both sides, is equal to the principal root of one
minus sine squared of Y. So, we could rewrite this as being equal
to one over, one over, instead of cosine of Y, we could rewrite it as one
minus sine squared of Y. Now why is this useful? Well, sine of Y is just X. So this is the same if we just substitute
back in, let me just write it that way so it's a little
bit clear. I could write it as sine Y squared. We know that this thing right over here is
X. So this is going to be equal to, and we
deserve a little bit of a drumroll. One over the square root of one minus,
instead of sin of Y, we know that X is equal to sin of Y. So, one minus X squared. And so, there you have it. The derivative with respect to X of the
inverse sine of X is equal to one over the square root of one minus X squared, so
let me just make that very clear. If you were to take the derivative with
respect to X of both sides of this, you get dy,dx is equal to this on
the right-hand side. Or we could say the derivative with
respect to X of the inverse sine of X is equal to one over the square root
of one minus X squared. Now you could always reprove this if your
memory starts to fail you, and actually, that is the best way to
really internalize this. But this is also just a good thing to
know, especially as we start, as we go into more and more
calculus and you see, you might see this in expression and you might say,
oh, okay, you know, that's the derivative of the inverse sine of X,
which might prove to be useful.