Differentiation of Inverse Functions: Example 4

by Batool Akmal

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    00:01 Let’s look at our last inverse function when it comes to trig. We are going to differentiate y equals to tan inverse of x. Now it’s very similar to the two inverse functions we’ve just differentiated.

    00:13 The only slight differences that we’re going to use a different identity to rearrange it at the end.

    00:18 But still, give it a go yourself and then I will follow and you can check. So, we have our function y equals to tan inverse of x. For an inverse function, we’re going to take the tan to the other side.

    00:33 So that becomes tan y = x. You can rewrite that as x = tan y. Then we differentiate so we can say that dx/dy and you know that tan of anything differentiates to sec² of y. Don’t forget that we’re looking at dy/dx, not dx/dy. So if we do dy/dx on this side, we have to change it to 1 over sec²y on the other side.

    01:04 Now, here’s the different identity that I was talking about. You were looking for something that connects tan and sec so that you can use that identity and rearrange it. You may remember that the identity that we can use is 1 + tan²y = sec²y. You don’t really have to rearrange anything here because you have sec² of y here and you know what that is equal to. So, we don’t have to rearrange.

    01:33 We can simply replace this with 1 over 1 + tan²y. Still in terms of y, but look at our very first line of the solution here where we said that x = tan y. So because you have two tans here or you have tan², you have x² there. Finally, you can rewrite this as dy/dx = 1 over 1 + x². Again, in your formula booklets or textbooks, you will be given this most of the time. So y equals to tan inverse of x differentiates dy/dx to 1 over 1 + x². Now, these results you’re not expected to learn because they’re usually given or alternatively if you really need to, you can derive. We’ll need to practice how to use these results with a function of a function. For example, if it was tan inverse of 5x, how do we do those questions? I’ll give you a chance to practice those in your exercise lecture before we go through them together. So, if you now have a look at your exercise lecture, you’ll see that there is a variety of different types of question. The first one just letting you practice dx/dy and to swap it around. The second one then involves an inverse function. But you can just go straight to the standard result rather than deriving it. Same for the third one, you can go straight or you can use the result of tan inverse of x and see if you can work that out.

    03:09 Remember there is a little bit of chain rule in those. But try it out and then we’ll do it together.

    About the Lecture

    The lecture Differentiation of Inverse Functions: Example 4 by Batool Akmal is from the course Differentiation of Inverse Functions.

    Included Quiz Questions

    1. secΘ = √(1+tan²Θ)
    2. secΘ = √(1 - tan²Θ)
    3. secΘ = √(1 + tanΘ)
    4. secΘ = √(1 - tanΘ)
    5. secΘ = 1 + tan²Θ
    1. dy/dx = 1 / (1 + x²)
    2. dy/dx = 1 / (1 - x²)
    3. dy/dx = -1 / (1 - x²)
    4. dy/dx = -1 / (1 + x²)
    5. dy/dx = 1 / √(1 - x²)
    1. dy/dx = 12x / (1 + x⁴)
    2. dy/dx = 12x / (1 + x²)
    3. dy/dx = 6 / (1 + x⁴)
    4. dy/dx = 12x / (1 - x²)
    5. dy/dx = -12x / (1 + x⁴)

    Author of lecture Differentiation of Inverse Functions: Example 4

     Batool Akmal

    Batool Akmal

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