# Differentiating Trigonometric Functions

by Batool Akmal

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DLM Differentiation of Trigonometric Functions Calculus Akmal.pdf
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00:01 Okay, here we're going to look at actually differentiating a function.

00:04 Now, this shouldn't be frightening because you already did this.

00:07 You've done the differential of sin, cos, and tan.

00:10 So anything after that is just going to be easy.

00:13 Let's have a look at the differential of cot of x and what we can do to differentiate it.

00:19 So, we are finding the derivative d by dx of cot of x.

00:25 Now, cot is a new identity for us. So, we don't really know what the answer to this is yet but you can break it down. So, we can write this as d/dx and instead of using cot, why don't we change it to 1 over tan x because we're more familiar with that.

00:43 So, using the identities at the start, we know that cot is 1 over tan and now we just have to differentiate 1 over tan x.

00:50 A couple of ways of doing this but observe that I have a function at the top and a function at the bottom. Hopefully, this is ringing some bells.

01:00 You have a function that is dividing a function, I have just called them u and v.

01:05 So, hopefully, we'll remember that we will have to use the quotient rule.

01:09 Quotient rule, dy/dx is vdudx minus udvdx over v squared.

01:17 In our case, our u is 1 and our v is tan x. When we differentiate u dashed, that's zero, that's always good news, and then when we differentiate tan we get sec squared x.

01:32 Let's apply the quotient rule now. So, we're doing dy/dx, vdudx, so we'll have tan of x multiplied by zero and then minus 1 times with sec squared of x over our v squared, so that's going to be tan x all squared. We're basically done but there are things we can do to simplify this to give us a nice answer.

02:01 So, we have minus sec squared of x at the top because the first term just becomes zero and then we have tan squared of x at the bottom.

02:10 Now instead of leaving it like this, let's just use some more identities.

02:14 So you know that sec from the start is cos, okay? So, if sec is cos, sec squared will be 1 over cos squared.

02:23 Apology, sec is a 1 over cos, so sec squared will be 1 over cos, all squared.

02:28 It's better if I write that down. So, we have 1 over cos x.

02:33 Remember, that it's squared because we're dealing with sec squared and then we also know that we were timesing this with 1 over tan squared x.

02:44 Tan is the same as sin over cos. So, if we write that here, we've got tan x is the same as sin x over cos x, one of the identities that we should know from previous use.

02:57 So, tan is sin over x. So, we can change this time to sin over x but remember that because it's 1 over tan squared, it would be cos over sin.

03:08 So if I did 1 over tan x that should be cos x over sin x. We just swap it.

03:14 So we're flipping the fraction. So if we do 1 over this, it becomes that.

03:19 Okay? So let me just rewrite that. So, I've got minus 1 over cos x, all squared and then if I multiply this with 1 over tan squared x, that should be cos x over sin x, all squared.

03:35 I hope you can see what's about to happen. Let me just do this in 2 steps.

03:40 So we have minus 1 when you square 1, you just get 1.

03:44 At the bottom, I'll have cos x squared, I have cos x squared when I square this and then I have sin x squared here. The cos x squared, cos x squared cancels out.

03:57 We now end up with minus 1 over sin squared x and this is done but you also can show your knowledge off a little bit more by using the fact that 1 over sin is cosec. So, you can write this as minus cosec squared x and if you go right to the beginning, you will see that we mentioned that the derivative of cot x is minus cosec squared x and I said that these are the kind of functions that you don't actually need to learn the derivative for because you can always just derive it. So our last example now.

04:33 We are differentiating 5 plus sin x divided by 2 plus cos x.

04:38 Quite obvious looking, it's a function being divided by a function.

04:43 So well talk about what rules we have to use.

04:45 You'll have to recall the methods that we've learned previously in order to differentiate this.

04:50 So we have a function y equals to 5 plus sin x over 2 plus cos x.

04:58 The easiest thing to do without having to rearrange or over complicate things is just to split it as u and v and obviously, from here you can see that in order to differentiate, we are going to have to use the quotient rule.

05:15 So, we know that the quotient rule is dy/dx is vdudx minus udvdx over v squared.

05:23 Let's put it all together here. So u equals to 5 plus sin x, v equals to 2 plus cos x.

05:32 Let's differentiate each one. 5 is a constant, so that goes away and sin x remember, differentiates to cos x. V dashed, 2 is a constant so that goes away, and cos x differentiates to minus sin x. So, nice and straightforward so far.

05:50 Let's put it all into dy/dx. Vdudx minus vdudx, so we can multiply cos x with 2 plus cos x, then we have a minus in the middle and then we have minus sin x multiplied by 5 plus sin x and it's all over v squared. So that's 2 plus cos x squared.

06:17 We are basically done with the differentiation, all we have to do is tidy this up.

06:22 So I can multiply cos through. So I've got 2 cos x plus cos squared x, leave the minus for now, this little minus here. Let's just expand this out first.

06:34 So if I leave the minus out, leave my brackets. So, I have minus 5 sin x and then I have minus sin squared x, all over 2 plus cos x, multiply this 2 now with the minus just to get rid of the brackets.

06:54 So we have 2 cos x, plus cos squared x, plus 5 sin x, and then plus sin squared x.

07:04 Okay I was hoping that this would happen, which is good news.

07:10 If you've observe here, you have a cos squared x plus sin squared x.

07:14 So if you remember previously, one of the identities says that cos squared x if I add this with this. Cos squared x plus sin squared x equals to 1.

07:23 So that now leaves us with 2 cos x plus 5 sin x. Again, this term here and this term here, adds together to give you 1. So we've got plus 1. Let me just write that here, the identity.

07:38 So sin squared x plus cos squared x equals to 1. So, when you write them next to each other, you'll be able to see that that goes to 1 and then it's all being divided by 2 cos x squared.

07:49 So the derivative of something this complicated is just this answer here and again, you can find your differential at different points by just substituting values in.

08:01 So now it's your turn to practice everything that we've learned.

08:05 It's now a chance for you to apply the differentials of sin, cos, and tan, and also all the other rules that we've learned previously.

08:12 So as you differentiate, don't forget the chain rule, don't forget the product rule, don't forget the quotient rule, and we're bringing it all together in this exercise. Good luck.

### About the Lecture

The lecture Differentiating Trigonometric Functions by Batool Akmal is from the course Differentiation of Trigonometric Functions.

### Included Quiz Questions

1. 5[sin²(x) + 2xsin(x)cos(x)]
2. [sin²(x) + 2xsin(x)cos(x)]
3. 5[sin²(x) - 2xsin(x)cos(x)]
4. -5[sin²(x) + 2xsin(x)cos(x)]
5. 5[cos²(x) + 2xsin(x)cos(x)]
1. 2tan(x)sec²(x)
2. 2tan(x)
3. tan(x)sec²(x)
4. 2sec²(x)
5. -2tan(x)sec²(x)
1. 0
2. cos²(x)/sin²(x) -1
3. -2
4. 1
5. sec²(x) - cot²(x) - 1
1. dy/dx = (cos(x) - sin(x) + 1)/(1-sin(x))²
2. dy/dx = (cos(x) + sin(x) - 1)/(1-sin(x))²
3. dy/dx = (cos(x) - sin(x) + sin²(x) - cos²(x))/(1-sin(x))²
4. dy/dx = (cos(x) - sin(x) - sin²(x) + cos²(x))/(1-sin(x))²
5. dy/dx = (cos(x) - sin(x) - 1)/(1-sin(x))²

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