00:01
So let's get straight into it now, we are now going to derive the answer for the gradient of sin x.
00:08
We already know the answer to this, I know.
00:10
So we've discussed already that sin x differentiate the cos of x
but we're going to prove it to ourselves, we're gonna develop our algebraic,
our proving, our analysis skills a little bit more
and we're hopefully going to be excited when we see the answer.
00:24
So back to the very first lecture of differentiation from first principles,
let me just remind you every time we're faced by a function that we want to prove,
we can always go back to the definition of the derivative.
00:36
So if I remind you, if you have, dy/dx of a function, you say limit delta x tends to zero
of f of x plus delta x minus f of x over delta x.
00:55
We are now trying to prove sin of x so we want the derivative of sin of x
as the limit of delta x tends to zero. Okay, so instead of using f,
we'll now use s, sin of x plus delta x minus the original sin x all over delta x.
01:20
Right, we can start using our addition law, so if you look at this function here,
this is basically sin a plus b. So you can compare this with sin a plus b
that we looked up previously, so your a is your x and b is your delta x.
01:41
So we can use that identity to expand this.
01:45
We said that sin a plus b is sin a cos b plus sin b cos a
but in this case we'll obviously do this with x's and deltas.
01:54
So we got sin x cos delta x plus sin delta x cos x, so that's this part expanded.
02:05
Don't forget to subtract it with sin x all divided by delta x.
02:12
And also don't forget that we still need to consider that the limit of delta x tends to zero.
02:17
Okay the next thing we can do is we can combine our sin x terms
so this term has a sin x and this term has a sin x and combine our cos x terms which is this one.
02:29
So if we put them together, if I rewrite this one's again, so it's a bit clearer,
I have sin x cos delta x and I'm gonna write sin x next to it so this term
just change its position and then I have sin delta x cos x
which is this term here by itself all over delta x.
02:50
Okay, this is what I've done this so that I can take a common factor out of these two terms of sin x.
02:58
So you can see they both have sin x. So I can take sin x out, leaving me with cos delta x
minus 1 over delta x and then in this term I can take a common factor of cos x
so we'll say cos x leaving us with sin delta x over delta x
and remember that for each of them we're still applying that the limit of delta x tends to zero
and the limit of delta x tends to zero. Okay, we're almost done now,
so sin x is the limit of delta x tends to zero isn't really affected so that change, stays as sin x.
03:44
We have cos delta x, so this value here you have to think now
what happens to cos as it approaches zero. So you can either use a circle or a calculator,
let me just remind you what the cos graph looks like.
03:59
Like so, this is zero, so add zero the cos graph is at 1.
04:04
So as this graph approaches zero, you can say that this value here approaches 1.
04:11
So at the top you have 1 minus 1 which gives you zero divided by anything is zero.
04:17
So you have sin x that's been multiplied by zero.
04:20
Okay, the next of it is cos x, obviously is an affected by the limit because there's no delta x here.
04:28
This is our proof that we just did earlier, so we said that sin x over x
as the limit of x goes to zero goes to 1 but instead we're now using delta x, delta x
and the limit of delta x, same thing. So we know that this entire thing approaches 1.
04:45
So we can now multiply this with 1 and finally our answer or the final answer
if we just write it here to make it a little bit more dramatic,
we can say that dy/dx as the limit of delta x times zero is simply cos x
because that term goes away. And so, worst case scenario
if you ever can't remember the derivative of sin x,
you know the whole methods to actually derive it.
05:09
Hopefully, you don't have to do this, this can just be to prove something
but you can learn that sin x differentiate to cos x and the differential of sin x is simply cos x.