Inverse Differentiation: Exercise 3

by Batool Akmal

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    00:01 The last exercise question, you’ll see that now, this has built up even a step further. We’ll talk about why that is but there’s quite a few little things happening in this question that we’ll have to deal with one at a time. So, if we look at this here, we’ve got y equals to 10x multiplied by tan inverse of (6x).

    00:22 The first thing I’m going to do is just write the standard result of y equals to tan inverse of (x) which you could just look up, gives you dy/dx = 1 over 1 + x². You can also derive this.

    00:40 If you’ve enjoyed doing that, you can just do that from scratch and derive this result and then use it on to this question. But back to this question, let’s look at the different components of it.

    00:50 First of all, we have a tan 10x, then we have a tan inverse of (6x). Within tan inverse of (6x) we have a function inside of a function. Basically, there are two things that are multiplying together.

    01:05 Then one of them has a function inside of a function. So, lots of different methods hopefully going through our minds at the moment. The first thing, when you’re differentiating two things that are timesing together or that are products, we have to use the product rule. The second thing, when you have a function inside of a function, we obviously have to use the chain rule.

    01:25 Let’s just go straight into it and see how we deal with this. We’re going to use the product rule.

    01:32 So, you have your two terms here. You have your u and then you have your v. U is 10x and v is tan inverse of (6x). Now usually, we do this much faster but because we have a few little things going on, we’ll do them step by step. I actually break it down into the formula. U dashed or du/dx will just give you 10. V dashed here, we’ll have to be a little bit careful about. Now, you know that tan inverse of (x), so when you differentiate tan inverse of (x), you get 1 over 1 + x². We are now looking at tan inverse of (6x). So you have a function, tan inverse. And then you have a function inside of tan inverse which is 6x. So, the first thing that you do when you apply the chain rule is you differentiate the outside function. So tan inverse of anything goes to 1 over 1 + the thing, so that’s (6x)². Then don’t forget to multiply it with the differential of the inside function, which is this function here which is just 6.

    02:39 If I just tidy this up a little bit, that gives me 6 over 1 + 36x² when I square the 6. We put this straight into the product rule. So, we can now say that dy/dx is vdu/dx + udv/dx or the other way around.

    03:00 With the product rule, it doesn’t matter. This gives you 10tan inverse of (6x) and then we plus.

    03:08 We now have to multiply these two functions together. So, we’ll have 10x multiplied by 6 over 1 + 36x².

    03:22 We’ve done the differentiation. We just make this a little bit neater so we can do tan inverse of (6x).

    03:29 These two numbers can multiply at the top. So you get 6dx over 1 + 36x². Instead of taking this as a common factor and multiplying it through, we can just leave it here because it’s nice and straightforward. At this point, I could take 1 + 36x² as my common factor.

    03:50 But that would just make things a little bit more complicated because anything that I do or if I multiply this 1 + 36x² with tan inverse of (6x) will have the term tan inverse of (6x) in it.

    04:03 So, I won’t really be able to add or combine any more terms. It might just be easier to leave it like this is unless you need to use it in a different form. In that case, you can times it through. But for now, I feel this is probably the easiest point to stop at and the most simplified this equation can be.

    About the Lecture

    The lecture Inverse Differentiation: Exercise 3 by Batool Akmal is from the course Differentiation of Inverse Functions.

    Included Quiz Questions

    1. dy/dx = 7 / (1 + x²)
    2. dy/dx = 7 / (1 - x²)
    3. dy/dx = 7 / √(1 - x²)
    4. dy/dx = -7 / (x² + 1)
    5. dy/dx = -7 / (1 - x²)
    1. dy/dx = 42x / (1 + 9x⁴)
    2. dy/dx = 42x / (3x² + 1)
    3. dy/dx = -42x / (1 + 9x⁴)
    4. dy/dx = 21x / (1 + 3x²)
    5. dy/dx = -21x / (1 + 9x⁴)
    1. dy/dx = 7tan⁻¹(3x) + 21x / (9x² + 1)
    2. dy/dx = 7tan⁻¹(3x) + 7x / (9x² + 1)
    3. dy/dx = 7tan⁻¹(3x) + 21x / (1 - 9x²) ]
    4. dy/dx = 7tan⁻¹(3x) + 7 x / (9x² - 1) }
    5. dy/dx = 7tan⁻¹(3x) + 7 x / (9x² + 3x)

    Author of lecture Inverse Differentiation: Exercise 3

     Batool Akmal

    Batool Akmal

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