### Video Transcript

Find the derivative of the inverse of tan π₯.

So, the first thing Iβm gonna do is let π of π₯ equal the inverse of tan π₯. So then, if I take the tan of both sides of the equation, Iβm gonna get tan of π of π₯ is equal to π₯. And thatβs because if you take tan of inverse tan of π₯, then this is just equal to π₯. Itβs also worth noting at this point something thatβs gonna be useful later on. And that is that the first derivative of π of π₯ must be equal to the first derivative of inverse tan π₯ because we already said that π of π₯ is equal to inverse tan π₯.

So now, if we say letβs look at the derivative of both sides of our equation, we can say that the derivative of tan π of π₯ is equal to the derivative of π₯. So now, what we can do is use one of our standard derivatives to help us work out what the derivative of tan π of π₯ is going to be. And thatβs because if we take the derivative of tan ππ₯, this is equal to π sec squared ππ₯. And this is where the π before the sec squared ππ₯ is just the result of finding the derivative of ππ₯. So therefore, itβs gonna be equal to sec squared π of π₯ multiplied by the derivative of π of π₯, and this is equal to one. And thatβs because the derivative of π₯ is just one.

And now, what we can do is substitute in the inverse of tan π₯ for π of π₯. Because we know from the beginning thatβs what theyβre equal to. So, therefore, what weβre gonna get is sec squared inverse of tan π₯ multiplied by the derivative of π of π₯ is equal to one. So, therefore, if we rearrange this, we can say that the derivative of π of π₯ is equal to one over sec squared inverse tan of π₯.

So now, what we can do is use one of our trig identities. And that one is that sec squared π₯ is equal to tan squared π₯ plus one. And if we substitute that in, so when we substitute this in, we get the derivative of π of π₯ is equal to one over tan squared the inverse of tan π₯ plus one. So, therefore, weβre gonna have the derivative of π of π₯ is gonna be equal to one over π₯ squared plus one. Thatβs cause tan squared of the inverse tan of π₯ gives us π₯ squared.

So, therefore, the derivative of the inverse of tan π₯ is gonna be equal to one over π₯ squared plus one.