Question Video: The Derivative of an Inverse Tangent Function | Nagwa Question Video: The Derivative of an Inverse Tangent Function | Nagwa

Question Video: The Derivative of an Inverse Tangent Function Mathematics • Higher Education

Find (d/dπ‘₯) tan⁻¹ π‘₯.

02:47

Video Transcript

Find the derivative of the inverse of tan π‘₯.

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

So now, if we say let’s look at the derivative of both sides of our equation, we can say that the derivative of tan 𝑓 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 𝑓 of π‘₯ is going to be. And that’s because if we take the derivative of tan π‘Žπ‘₯, this is equal to π‘Ž sec squared π‘Žπ‘₯. And this is where the π‘Ž before the sec squared π‘Žπ‘₯ is just the result of finding the derivative of π‘Žπ‘₯. So therefore, it’s gonna be equal to sec squared 𝑓 of π‘₯ multiplied by the derivative of 𝑓 of π‘₯, and this is equal to one. And that’s because the derivative of π‘₯ is just one.

And now, what we can do is substitute in the inverse of tan π‘₯ for 𝑓 of π‘₯. Because we know from the beginning that’s what they’re equal to. So, therefore, what we’re gonna get is sec squared inverse of tan π‘₯ multiplied by the derivative of 𝑓 of π‘₯ is equal to one. So, therefore, if we rearrange this, we can say that the derivative of 𝑓 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 𝑓 of π‘₯ is equal to one over tan squared the inverse of tan π‘₯ plus one. So, therefore, we’re gonna have the derivative of 𝑓 of π‘₯ is gonna be equal to one over π‘₯ squared plus one. That’s 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.

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