# Differentiation of Trigonometric Functions: Exercise 4

by Batool Akmal

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00:01 So the last question, which is a little bit of an algebraic challenge, is 2 functions dividing each other but this time there's no numbers, we're just dealing with letters.

00:10 The letters are constant so a and b could be any numbers and we're just deriving a general formula or a general result.

00:18 So, we have y equals to a plus b sin of x over a minus b sin of x.

00:28 Look at your function. What is happening' Two functions are dividing.

00:34 When you are to differentiate something that's dividing, you will use the quotient rule.

00:39 So, these things should be coming more naturally to us. This is our u, this is our v.

00:44 Our quotient rule is this, dy/dx is vdu/dx minus udv/dx over v squared.

00:52 Let's break it down into its u's and v's. So u equals to a plus b sin x, v equals to a minus b sin x.

01:01 Differentiate it, u dashed. Now, here's the thing, a is any constant, it's any number so it will just disappear so don't just keep it there because it's a letter, a is a number so it disappears, b is a constant so you can imagine it's 5 sin x and you differentiate it. So b stays and sin x differentiates to cos x. It's not a function of a function, no chain rule, just straight forward, differentiate sin x to cos x. Similarly for v dashed, so let's just stay with the same notation rather than dv/dx.

01:34 The a disappears and then b or the sin x goes to cos x, so this becomes minus b cos of x.

01:42 Okay, let's put these all into our formula for our quotient rule, vdu/dx minus udv/dx.

01:52 So we can now say that dy/dx is b cos x a minus b sin x, doesn't matter what order I multiply them in, minus I have minus b cos x multiplied by a plus b sin x. So this is all being divided by v squared, so our v is a minus b sin of x and it's all squared. Okay, so the differentiation part is done.

02:25 All we have to do is simplify this out. So, I'll do that by multiplying this through and then multiplying this through and then seeing if we can put them together to make it look a bit tidier. So b cos x multiplied by a will give me ab cos x, b cos x multiplied by minus b sin x will give me minus b squared sin x cos x.

02:52 Let me just leave the minus here for a moment and just solve the inside of the brackets.

02:57 So that gives me minus ab cos x and then I have minus b squared sin x cos x, all over a minus b sin x squared. Okay, when we times it through with the minus, so this and this, you'll see that the signs will change.

03:20 So this will become a plus and that will also become a plus, so let's just do that.

03:25 And then hopefully we can cancel some things and add things, cos x minus b squared sin x cos x plus ab cos x and then you have a minus and a minus to give you plus b squared sin x cos x and this is still all over a minus b sin x all squared.

03:51 So we can see that we have a minus b squared sin x cos x and a plus b squared sin x cos x.

03:59 We now just have ab cos x here. We now just have ab cos x here, plus another ab cos x.

04:07 So, this gives you 2ab cos x and we are dividing it with the same denominator that stayed with us for a while, a minus b sin x all squared.

04:23 So, by using a little bit of algebra, we've simplified our derivative and got 2ab cos x divided by a minus b sin x all squared.

The lecture Differentiation of Trigonometric Functions: Exercise 4 by Batool Akmal is from the course Differentiation of Trigonometric Functions.

### Included Quiz Questions

1. 5cos(x) / [2 - sin(x)]²
2. 5cos(x) / [2 + sin(x)]²
3. cos(x) / [2 - sin(x)]²
4. cos(x) / [2 + sin(x)]²
5. 5cos(x).[2 - sin(x)]²
1. sin²(x)
2. cos²(x)
3. cot²(x)
4. 1
5. tan²(x)
1. cot(x)
2. cot²(x)
3. tan(x)
4. tan²(x)
5. cosec(x)

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