So if you enjoyed the proof for sin x,
now is your chance to take the lead on the proof for cos x.
It's very similar to what we've just done with the same methods.
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.
So, using exactly the same method, let's now find the derivative for cos x.
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.
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.
So, that's what we're trying to do as the limit of delta x tends to zero.
So, we can rewrite this as cos x plus delta x minus the original cos x all over delta x.
This functional now, we will need to expand using our double angle
or addition law formulas that we gave right at the start.
So, remember that you can compare this with cos of a plus b.
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.
So, if we expand this using our rule, we can say cos x cos delta x.
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.
So you have a cos x value here and a cos x value here.
So we can just rewrite it in order to write them next to each other so it doesn't look too confusing.
So we have cos x delta x minus cos x and then you also have minus sin x sin delta x.
And don't forget that this is all over delta x.
The reason we've done that is so we can rewrite it as a common factor.
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.
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.
So for each individual section of this equation, we now apply delta x equals to zero.
Let's do that. Cos x isn't really affected because there's no delta x there. So, we have cos x.
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.
So, 1 minus 1 gives you zero. So this entire term becomes zero.
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.
Our x in this case is delta x, so it's the same thing, it's still 1.
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.
So we said at the beginning that cos of x equals to minus sin x and we've proven now.