# Differentiation of Trigonometric Functions: Exercise 2

by Batool Akmal

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00:01 Our next example is now asking us to differentiate 1 over sin x.

00:05 Now think if you can remember what 1 over sin x was called.

00:08 It's cosec of x. But we like to deal with things written in terms of sin, cos, and tan, so rather than overcomplicating it, let's just start differentiating this.

00:18 So I have y equals to 1 over sin x. If you identified or recognized that this is the same as cosec, that's brilliant, you're obviously learning your identities.

00:28 But in this case, we can see that we have a number at the top and a number at the bottom.

00:32 You can take you entire sin x up and then just treat it like the chain rule.

00:38 So by that I mean you could do this, sin x to the minus 1 and then it's very similar to the previous example, so you could use it as the chain rule.

00:46 But the idea is to practice the quotient rule so I won't do that.

00:49 We'll just rewrite this as the quotient of u and v, so I'll say the top value is u and the bottom value is v.

00:57 The quotient rule, dy/dx is vdu/dx minus udv/dx over v squared.

01:06 So we separate it out and then we put it into the formula. So, u equals to 1, v equals to sin of x, u dashed is zero and v dashed is cos of x. Go straight into the formula, we're using vdu/dx minus that so dy/dx is just going to be sin of x multiplied by zero minus 1 times cos of x and then you divide it by the squared which is just sin x squared or sin squared x. Sin multiplied by zero just gives you zero, that just goes away, leaving you with minus cos of x over sin x squared. This is done.

01:54 There are lots of other things you could do with this, so you could write this as minus cos x and then your sin x you could write us, let's just do it like this, 1 over sin x all squared for now.

02:08 You can see that there are too lots of sin x, so you can write this as minus cos x over sin x times sin x or sin squared x. You can take 1 bit of cos and sin and if you remember from the previous lecture, we said that tan x is sin over cos, so we also said that cos x over sin x is just 1 over tan. So, it's just inverted.

02:39 So, here you can see that we have cos over sin which is just 1 over tan.

02:44 So, we can rewrite this as minus 1 over tan x and then you still have sin x at the bottom.

02:52 We know that 1 over tan x is the same as cot so that's minus cot x and then we still are timesing it with 1 over sin x and use this knowledge that we did at the start, sin x is the same as cos x so you can rewrite this as minus cot x cosec of x.

03:16 And what we've just done is found the derivative for cosec x, so if you, like we said earlier, you could write 1 over sin as cosec of x and if someone asks you to differentiate that, you'd get this answer similar to someone asking you to differentiate this because they're both the same thing.

03:35 So, we've worked out that the derivative of cosec of x is minus cot x cosec x.

The lecture Differentiation of Trigonometric Functions: Exercise 2 by Batool Akmal is from the course Differentiation of Trigonometric Functions.

### Included Quiz Questions

1. dy/dx = -12cot(6x)/sin²(6x)
2. dy/dx = 12cot(6x)/sin³(6x)
3. dy/dx = -2cot(6x)/sin²(6x)
4. dy/dx = -12cot(6x)/sin³(6x)
5. dy/dx = +6cot(6x)/sin²(6x)
1. dy/dx = sec(x).tan(x)
2. dy/dx = cosec(x).tan(x)
3. dy/dx = sec²(x).tan(x)
4. dy/dx = sec(x).cot(x)
5. dy/dx = -sec(x).tan(x)

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