So let's get straight into it now, we are now going to derive the answer for the gradient of sin x.
We already know the answer to this, I know.
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.
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.
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.
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.
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.
So we can use that identity to expand this.
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.
So we got sin x cos delta x plus sin delta x cos x, so that's this part expanded.
Don't forget to subtract it with sin x all divided by delta x.
And also don't forget that we still need to consider that the limit of delta x tends to zero.
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.
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.
Okay, this is what I've done this so that I can take a common factor out of these two terms of sin x.
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.
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.
Like so, this is zero, so add zero the cos graph is at 1.
So as this graph approaches zero, you can say that this value here approaches 1.
So at the top you have 1 minus 1 which gives you zero divided by anything is zero.
So you have sin x that's been multiplied by zero.
Okay, the next of it is cos x, obviously is an affected by the limit because there's no delta x here.
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.
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.
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.