Differentiation of Trigonometric Functions: Exercise 3

by Batool Akmal

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    00:01 Let's look at our next example. We are now looking at a function x multiplied by cos squared of x. So we can rewrite this and just kind of talk about what function this is, so we have cos squared of x.

    00:14 There are a couple of things we can do to make it easier and more relevant to what we're doing.

    00:19 You could rewrite this as x and you can write this as cos x squared because cos squared of x is the same as cos x all squared. Right. Look at your function.

    00:31 Decide what it is before we do anything else. So, when you observe, you'll see that you have an x function that is being multiplied with a cos x squared function.

    00:42 Hopefully, things are going through your mind, chain rule, product rule, quotient rule, which one to choose. This is a product they're timesing together, so we're going to use the product rule. The product rule states that dy/dx is vdu/dx plus udv/dx. You can split this into your u and v, doesn't matter which one's which, so u is x and v equals to cos x squared.

    01:07 If you differentiate u, you will get 1. When you differentiate this, use the chain rule, so bring the power down, decrease the power by 1 and multiply it with the differential of the inside. So cos x differentiates to minus sin x.

    01:24 Tidy this up, this gives you minus 2 cos x sin x. Right. Let's put this into our formula.

    01:35 So we can now say that dy/dx is vdu/dx, so if we just do it in this order, you will get cos x squared multiplied by 1 and then you have plus x multiplied by minus 2 cos x sin x.

    01:52 This then gives you cos x all squared minus 2x cos x sin x.

    02:01 And there are things that you could do like take a factor of cos x.

    02:06 I would if you would like to, doesn't make too much of a difference, so you can take a factor of cos x out leaving you with cos x minus 2x sin x.

    02:17 And that's as far as we can simplify. Later on, you will learn more identities to differentiate things like this. So, you could use the squared identities, the double angle identities to actually rearrange and differentiate, but for now, this is as simple as we can make this answer.

    About the Lecture

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

    Included Quiz Questions

    1. dy/dx = 18xcos(x² + 1) - 18x³sin(x² + 1)
    2. dy/dx = 9xcos(x² + 1) - 9x³sin(x² + 1)
    3. dy/dx = 18xcos(x² + 1) - 18x²sin(x² + 1)
    4. dy/dx = - 18x³sin(x² + 1)
    5. dy/dx = -9xcos(x² + 1) - 9x³sin(x² + 1)
    1. dy/dx = 15x²sin²(x) + 10x³sin(x)cos(x)
    2. dy/dx = 5x²sin²(x) + 5x³sin(x)cos(x)
    3. dy/dx = 5x²sin²(x) - 5x³sin(x)cos(x)
    4. dy/dx = 5x²sin²(x) + 10x³sin(x)cos(x)
    5. dy/dx = 15x²sin²(x) + 10x³sin(x)

    Author of lecture Differentiation of Trigonometric Functions: Exercise 3

     Batool Akmal

    Batool Akmal

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