At this point, we've learned a ton of different rules for finding derivatives, and we know how to find the derivatives of basic trigonometric functions like sine or cosine. But we'll also need to know how to find the derivatives of inverse trigonometric functions, and that's exactly what we're going to start looking at here, focusing specifically on the derivatives of inverse sine and inverse cosine. Now we're going to walk through a familiar process to actually find the derivative of the inverse sine function to hopefully give you a bit of intuition as to why these rules are what they are. But once we have these rules, we can then use them to actually find the derivatives of some functions. So let's go ahead and jump into things here.
Now if y is equal to the inverse sine of x, that means that x is then equal to the sine of y. Since these are equivalent statements and we know how to find the derivative of sine, we can then use implicit differentiation to ultimately find the derivative that we're looking for of the inverse sine. So coming down here, having rewritten this as x equals sine of y, remember that based on our inverse sine function, y is restricted by these values. It has to be between negative pi over 2 and positive pi over 2. But now that we have this x equals sine y, let's start our process of implicit differentiation.
So we need to start here by taking the derivative d/dx on each side. Now we know that the derivative of x with respect to x is just 1. So then moving over to this right-hand side, the derivative with respect to x of the sine of y, I know that the derivative of sine is cosine, so this gives me cosine of y. Then having to apply the chain rule here because we're taking the derivative with respect to x. So we need to multiply here by the derivative of y with respect to x. That's dy/dx. Now this dy/dx is ultimately the derivative that we want to find. So we need to get that all by itself, which we can do by dividing both sides by the cosine of y in order to get that to cancel. Doing some rearranging here, this then gives me that dy/dx is equal to 1 over the cosine of y. So I've successfully found dy/dx.
But when we're working with derivatives with respect to x, we typically want our answer in terms of x rather than in terms of y. So in order to get this in terms of x, we're going to look to our trigonometric identities. Remember that sine squared plus cosine squared is equal to 1. So if I rearrange this, solving for the cosine of y, this then gives me that the cosine of y is equal to plus or minus the square root of 1 minus the sine squared of y. So I can plug this back into my equation over here.
This gives me over a plus or minus the square root of minus sine squared y. Now based on our original restrictions on y, this plus or minus square root is not going to work because we do not want the negative square root there. So we're only going to work with the positive square root. This is a positive value. Now, you may be thinking this is still in terms of y, so we have the same exact problem here.
But remember that we can rewrite the sine squared of y as sine of y squared. And what is the sine of y? The sine of y is just x. That's what we were working with from the very beginning. So I can just replace that sine y with x to ultimately give me that the derivative of the inverse sine of x, that's dy/dx, is equal to 1 over the square root of 1 minus x squared, having just replaced that sine of y with the x that I know it is.
So this is our rule for finding the derivative of the inverse sine function. Now there is a restriction on our x value here based on this being the inverse sine. The absolute value of x must be less than 1. Remember that with these restrictions, we don't want to think of them as another thing that we have to memorize and stress over. We want to focus on our rule because those restrictions follow just from our knowledge of functions.
So now that we have this rule, let's actually apply it to finding the derivative of a function. Here, we want to find the derivative of f(x) is equal to the inverse sine of 3x + 2. So we want to find here f'(x), our derivative. Now since here, I'm taking the inverse sine of 3x + 2 and not just x. That means I'm going to have to apply the chain rule.
But starting from my outermost function, that's the inverse sine. Based on the rule that we just found, we know that this derivative is equal to 1 over the square root of 1 minus in this case, it was x squared. But because here I'm taking the inverse sine of 3x + 2, that's what I need to plug in for x here. So this is 1 minus (3x + 2)^2. And then using the chain rule here, multiplying by the derivative of 3x + 2, which is just 3.
Now I can ultimately just write this 3 on the top of my fraction here to give me my final derivative here, f'(x). Now we ultimately found the derivative of the inverse sine using implicit differentiation, and we could follow that same exact process in order to determine the derivative of the inverse cosine. Now if we were to follow that process, we would find that the derivative of the inverse cosine is equal to negative one over the square root of 1 minus x squared, so the exact same thing that we got for sine except negative. Now we have that same exact restriction on x that its absolute value must be less than 1. So with this rule for finding the derivative of inverse cosine, let's work through one final example here and find the derivative of this function, g(x).
This is 4 times the inverse cosine of 6x. So finding our derivative here, g'(x), since I have this constant 4, I can just pull that out front and focus on the derivative of the inverse cosine of 6x. Now again, here, we're not just taking the inverse cosine of x, but 6x. So we need to attack this the same exact way we did in example a. So starting with our outermost function, the inverse cosine, we know based on the rule that we just learned that this derivative is equal or this derivative is equal to negative one over the square root of 1 minus, in this case, 6x because that's my inside function, so 6x squared.
Then applying my chain rule here, I need to multiply by the derivative of that inside function 6x, x, which is just 6. So I fully found my derivative here, but we can simplify this a little bit. If I go ahead and multiply that 4 times negative one times 6, that gives me negative 24 on the top. And then on the bottom, the square root of 1 minus 36 x squared, going ahead and squaring those values. So this gives me my final derivative here, g'(x).
Now as we saw for our basic trigonometric functions, we're also going to want to memorize these derivatives of our inverse trigonometric functions because they're going to come up a lot as we take more and more derivatives. But if you ever forget them, we know the process to figuring these out, and you can always look back to implicit differentiation to determine what these derivatives are. Now we're going to get some more practice using these derivatives of inverse sine and cosine coming up next. I'll see you there.
