If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains ***.kastatic.org** and ***.kasandbox.org** are unblocked.

Main content

Current time:0:00Total duration:4:56

AP.CALC:

FUN‑3 (EU)

, FUN‑3.E (LO)

, FUN‑3.E.2 (EK)

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.