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Derivative of inverse tangent

AP.CALC:
FUN‑3 (EU)
,
FUN‑3.E (LO)
,
FUN‑3.E.2 (EK)

Video transcript

we already know that the derivative with respect to X of tangent of X is equal to the secant of x squared which is of course the same thing as 1 over cosine of x squared now what we want to do in this video like we've done in the last few videos is figure out what the derivative of the inverse function of the tangent of X is or in particular let's see if we can figure out what the derivative with respect to X of the inverse tangent of X is and I encourage you to pause this video and use a technique similar to 1 or very close to the one that we've used in the last few videos to figure out what this is well let's set y equal to the inverse tangent of X Y is equal to inverse tangent of X that is the same thing as saying that the tangent of Y the tangent of Y is equal to X all I've done now you can kind of think of it as I've just taken the tangent of both sides right over here and now we can take the derivative of both sides with respect the derivative of both sides with respect to X and on the left hand side we can just apply the chain rule derivative of tangent of Y with respect to Y is going to be secant squared of Y which is the same thing as 1 over cosine of Y squared I like to write it this way it's keeps it a little bit simpler at least in in my brain but when we're applying the chain rule it's going to be the derivative of tangent of Y with respect to Y times the derivative of Y with respect to x times the derivative of Y with respect to X and on the right hand side the derivative of X with respect to X well that's just going to be equal to 1 and so we get if we want to solve for the derivative of Y with respect to X we just multiply both sides times the cosine of Y squared and we get the derivative of Y with respect to X is equal to a sign of y-squared and like we've seen in previous videos this isn't that satisfying because I've written the derivative of Y with respect to X as a as a function of Y but what we're really interested is writing it as a function of X and to do that we need to express this somehow in terms of the tangent of Y and the reason why the tangent of Y is interesting is because we already know that tangent of Y is equal to X so we can rewrite this using a little bit of trigonometric identities then we can with tangent of Y we can substitute all the tangent of y's with an X so let's see if we can do that and this seems a little bit tricky to introduce a tangent of Y we'd want to have a sine divided by a cosine that's what tangent is and this is just a straight up cosine squared Y so this is really going to take a little bit more experimentation then at least some of the other the last few examples we've done so one thing we could do is we could say hey let's just divide by 1 dividing by 1 never hurt anyone so we could say that this is the same thing this is the same thing as cosine squared Y and I'm really doing this too to see if I can start to express it as some type of a rational rational expression which might involve at some point a sine divided by cosine and I could have a tangent so let's divide by 1 but we know from the Pythagorean the Pythagorean identity that one is equal to sine squared of Y plus cosine squared of Y and so let's write that or we could write cosine squared of Y plus sine squared of Y once again why was I able to divide by this expression well this expression by the Pythagorean identity just really comes out of the unit circle definition of trig functions this is equal to 1 so I have not changed the value of this expression now what makes this interesting is if I want to introduce a sine divided by a cosine I could just divide the numerator and the denominator by cosine squared so let's do that let's multiply 1 over cosine or let's divide the numerator by cosine squared of Y and divide the denominator by cosine squared of Y or multiply each of them by 1 over cosine squared of Y and what's that going to give us well the numerator these character we're going to cancel you're just going to have a 1 and in the denominator this times this that's just going to be equal to 1 and then you're going to have sine squared sine squared Y / / cosine squared Y and this is this is the goal that I was trying to achieve I have a sine divided by cosine squared so this right over here this is the same thing actually let me just write it this way let me let me write it this way this is the same thing as sine of Y over cosine of Y cosine of Y whole thing squared which is of course the same thing as 1 over 1 + tangent of y tangent of Y squared this is equal to this now why is that useful well we know that X is equal to tangent of Y so this is going to be equal to this is going to be equal to 1 1 over 1 plus tangent of Y is equal to X x squared x squared which is pretty exciting we just figure out the derivative of Y with respect to X so the derivative of this thing with respect to X is 1 over 1 plus x squared so we could write that right up here so this is going to be equal to 1 over 1 plus x squared and we are done
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