Derivatives of inverse trigonometric functions
Derivative of inverse sine
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.