Lectures

Differentiation of Inverse Functions: Example 1

by Batool Akmal
(1)

Questions about the lecture
My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      DLM Differentiation of Inverse Functions Calculus Akmal.pdf
    • PDF
      Download Lecture Overview
    Report mistake
    Transcript

    00:01 We’ve just finished parametric equations and parametric differentiations. Remember before that, we did implicit differentiation. We looked before that at chain rule, quotient rule, and the product rule and we learned how to differentiate from first principles. We’re now going to look at this last section of differentiation where we’re looking at inverse differentiation. Now basically, that means it will teach us a technique of how to differentiate inverse functions such as sin inverse of x, tan inverse of x, and cos inverse of x. But before we do that, we’re going to talk a little bit about what to do if you have a function x in terms of y and then we’ll move forward to the actual inverse differentiation. If we look at our first example, we have a function where we’re being asked to find dy/dx. But if you’ll look at this closely, you’ll see that this is given in terms of x equals to a function of y. Now, we haven’t seen something like this before.

    01:04 We’ve done y equals to a function of x. We’ve done x and y as functions of t.

    01:10 But we haven't really seen a function of x in terms of y. Now usually, there are two things that we do.

    01:17 We might rearrange for y if it’s easy enough or we could do implicit differentiation.

    01:24 But if you look at this question, it’s a little bit more complicated. We have x in terms of a more complex-looking function of y. We ask ourselves whether there’s any way we can just make this a little bit easier and find an easier way to differentiate it. Let’s have a look at a technique that we can use to differentiate this function. Given a function, x = y² + 1 and y² – 1, now we said that there are two things that you could have done. You can make y the subject of the equation which then in this case will be a little bit complicated. You can also take this entire expression to the other side and then use implicit differentiation which would be slightly easier perhaps. But there is another way of doing this.

    02:15 That is by just keeping your objectives of finding dy/dx. Remember that what you want to do is calculate dy/dx.

    02:24 What you could do because you have an x function here is that you could calculate dx/dy, You could differentiate x in terms of y. Then remember to find your gradient, all you’d have to do is flip these functions. You would kind of swap the places to change it back to dy/dx. It’s quite a simple idea but it really helps with more complicated functions like these. What I’m saying here is instead of you having to rearrange, we could just momentarily say that we are going to differentiate x with respect to y. Remember, we always differentiate y with respect to x but we can differentiate x with respect to y with the condition that at the end, we swap them. So we changed, bring dy back to the top and dx to the bottom so that it becomes the gradient. So remember the gradient is this term here, it’s dy/dx. That’s what we’re going to be looking at doing. We want to find dx/dy.

    03:29 In order to do that, if you look at your function, it’s a function that’s being divided by another function.

    03:36 So, we need to use the quotient rule. Let’s just remind ourselves what the quotient rule is.

    03:41 The quotient rule states that in order to differentiate dy/dx, we can do vdu/dx – udv/dx all over v².

    03:53 This is the rule. We’re obviously not doing dy/dx yet but this is the rule that we use for the quotient rule.

    04:00 So, we pick our u and we pick our v. Our u in this case is going to be y² + 1 and the v will be y² – 1.

    04:11 We differentiate each one of them so we have u dashed. Now in terms of y will give you 2y.

    04:18 Then we have v dashed is also going to be 2y. Looking at this now, we now apply the quotient rule.

    04:26 So, we're going to apply dy/dx. We’re just looking at this definition here. We’re going to say vdu/dx, so we’re multiplying this expression with this. So we have y² – 1 multiplied by 2y – udv/dx here.

    04:48 We’re going to have 2y multiplied by y² + 1 and then we divide by v², (y² – 1)². We are going to tidy this up.

    05:03 So if we bring that back up here, let’s make this a little bit tidier. So this equals to 2y³ – 2y.

    05:15 and then we have another -2y³ – 2y all over (y² – 1)². You’ll see here that 2y³, 2y³ cancels out because one's a positive and the other’s a negative. So, we end up with -4y over (y² – 1)². Remember what this is.

    05:43 Initially, we started off by calling this dx/dy and we’ve used the formula of dy/dx.

    05:51 If I just relabel this here and put that as dx/dy because basically we’ve just differentiated this function here.

    06:00 We’ve taken the x and we’ve differentiated dx/dy. I hope this notation in between of the quotient rule didn’t quite confuse you. So, we’ve got our answer here but remember this is the answer for dx/dy.

    06:14 To find dy/dx, all you have to do is swap places. So the dy goes to the top and dx comes to the bottom.

    06:26 So we have to do the same on this side of the equation. We also have to swap this fraction.

    06:31 So we’ll leave the minus as it is. We’ve got (y² – 1)² over 4y. Here, you have your gradient dy/dx in terms of y just by using the fact that we can change dx/dy to dy/dx.


    About the Lecture

    The lecture Differentiation of Inverse Functions: Example 1 by Batool Akmal is from the course Differentiation of Inverse Functions.


    Author of lecture Differentiation of Inverse Functions: Example 1

     Batool Akmal

    Batool Akmal


    Customer reviews

    (1)
    5,0 of 5 stars
    5 Stars
    5
    4 Stars
    0
    3 Stars
    0
    2 Stars
    0
    1  Star
    0