# Derivative of cos(x)

by Batool Akmal

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00:01 So if you enjoyed the proof for sin x, now is your chance to take the lead on the proof for cos x.

00:07 It's very similar to what we've just done with the same methods.

00:10 It's incredible that we've learnt these derivatives all through school but we've never really seen where they come from and now is your chance to actually prove it to yourselves.

00:19 So, using exactly the same method, let's now find the derivative for cos x.

00:25 Remember, we kind of, not kind of we do know the answer to this already but let's see if we can actually derive it algebraically.

00:32 So, once again using differentiation from first principles, we are saying that dy/dx as the limit of delta x tends to zero is f of x plus delta x minus f of x divided by delta x. This time, we are differentiating cos x or we're finding what the derivative of cos x is.

00:54 So, that's what we're trying to do as the limit of delta x tends to zero.

00:59 So, we can rewrite this as cos x plus delta x minus the original cos x all over delta x.

01:09 This functional now, we will need to expand using our double angle or addition law formulas that we gave right at the start.

01:19 So, remember that you can compare this with cos of a plus b.

01:24 So this is the same as cos of a plus b and you can look that rule up and actually use that to expand cos x plus delta x.

01:33 So, if we expand this using our rule, we can say cos x cos delta x.

01:39 This time we have a minus sin x sin delta x minus cos x over delta x is our differentiation from first principles. Again, we can now combine our cos x terms.

01:56 So you have a cos x value here and a cos x value here.

01:59 So we can just rewrite it in order to write them next to each other so it doesn't look too confusing.

02:04 So we have cos x delta x minus cos x and then you also have minus sin x sin delta x.

02:13 And don't forget that this is all over delta x.

02:19 The reason we've done that is so we can rewrite it as a common factor.

02:24 So you can take cos x out of the first 2 terms. So, those 2 terms there, leaving you with cos of delta x minus 1 over delta x, and then you can take sin x out, leaving you with sin delta x over delta x. And again, don't forget, right at the start we said that the limit of delta x should tend to zero.

02:53 So for each part we can now apply it to each individual, let's just put delta x here so we can follow what we're saying.

03:04 So for each individual section of this equation, we now apply delta x equals to zero.

03:09 Let's do that. Cos x isn't really affected because there's no delta x there. So, we have cos x.

03:14 This if you remember from the previous proof that cos of delta x as delta x tends to zero, this goes to 1. Okay? Because cos of, as you get closer and closer to zero is 1.

03:27 So, 1 minus 1 gives you zero. So this entire term becomes zero.

03:31 Here, we have minus sin x which isn't affected by delta x and remember again, from our very first statement on one of the properties that we mentioned earlier, that sin x over x as the limit of x tends to zero is 1.

03:46 Our x in this case is delta x, so it's the same thing, it's still 1.

03:50 So that multiples with 1 and eventually, we can now say that dy/dx of this function is minus sin x and here we have derived the differential or the standard differential answer to cos of x.

04:06 So we said at the beginning that cos of x equals to minus sin x and we've proven now.

The lecture Derivative of cos(x) by Batool Akmal is from the course Differentiation of Trigonometric Functions.

### Included Quiz Questions

1. -sin(x)
2. sin(x)
3. -cos(x)
4. tan(x)
1. Cos(x)Cos(y) - Sin(x)Sin(y)
2. Cos(x)Cos(y) + Sin(x)Sin(y)
3. Sin(x)Cos(y) + Cos(x)Sin(y)
4. Sin(x)Cos(y) - Cos(x)Sin(y)
5. Cos²(x) + Cos²(y)
1. Lim x-->0 f(x) + Lim x-->0 g(x)
2. Lim x-->0 f(x) - Lim x-->0 g(x)
3. Lim x-->0 f(x) × Lim x-->0 g(x)
4. Lim x-->0 f(x) ÷ Lim x-->0 g(x)

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