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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 what 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 sine inverse of X is but we do know what the Sun 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 right over here this right over here is our goal we essentially want to figure out 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 were a little bit more familiar with and now we can do a little bit of implicit differentiation we could take the derivative of both sides with respect to X so the derivative of the left-hand side with respect to X and the derivative of the right-hand side with respect to X well 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 x dy/dx times dy/dx and the right-hand side what's the derivative of X with respect to X well that's obviously just going to be equal to 1 and so we could solve for dy/dx divided both sides by cosine of Y and we get the derivative of Y with respect to X is equal to 1 over cosine of Y now this still isn't that satisfying because I have the derivative in terms of why 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 and 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 1 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 1 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 1 minus sine squared of Y so we can rewrite this as being equal to 1 over 1 over instead of cosine of Y we could rewrite it as 1 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 and let me just write it that way so it's a little bit clearer I could write it assign 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 1 over the square root of 1 minus instead of sine of Y we know that X is equal to sine of Y so 1 minus x squared and so there you have it the derivative with respect to X of the inverse sine of X is equal to is equal to 1 over the square root of 1 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 would 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 you might say oh okay you know that's the derivative of the inverse sine of X which might prove to be useful