And now, the last one. Let's just prove the answer or the differential of tan x.
We know that we should be getting minus sec squared x
but let's try it out using exactly the same method.
So, if you enjoyed the previous two proofs,
I recommend that you try it yourself and then just catch up and check with us once you're done.
Okay, so we're going to derive the answer for tan of x.
Again, we're going to use differentiation from first principles or the definition of a derivative.
We're saying as the limit of delta x tends to zero is f of x plus delta x minus the original function,
all over delta x. This time, our function is tan of x. So, we are differentiating that.
Also remember, eventually we're going to take the limit of delta x to zero.
So we can rewrite this as tan of x plus delta x,
minus the original function tan of x all over delta x.
So we're going to have to expand this using the addition law for tan.
That's a little bit more complicated than the others
but we'll just have to use a little bit of algebra to tweak it and make it a little bit tidier.
So, if you recall that we can compare this with the tan of A plus B.
So our A now is x and the B is delta x, and you just apply the same formula.
Looking back at the formula, we find out that this expands to tan x not A.
So, tan x plus tan delta x over 1 minus tan x tan delta x.
Don't forget, we have to take tan x away from it as well,
and that it's all being divided by delta x and that the limit of delta x is tending to zero.
So all of those things still count. Right, let's look at just the top half first,
just because it's easier. So, let's try and simplify this. We can add these 2 fractions.
So we can treat this as 1 fraction, and this as 1 fraction.
So, we're going to add them up by taking a common denominator.
So I'm just gonna focus at the top part firstly.
So, we can say that we have tan x plus tan delta x.
I want to take my common denominator as 1 minus tan x tan delta x.
So in order to do that, I need to multiply this with 1 minus tan x tan delta x.
So I'll have minus tan x and I'm going to multiply it with this part here.
So, I've got 1 minus tan x tan of delta x.
So, I'm just trying to solve the top half of this equation.
I will bring this delta x in very shortly. So if I just expand this top bit,
so I've got the brackets that I need to open up.
So I've got tan x plus tan delta x and then I've got minus tan x.
When I multiply those 2 and then when I multiply that that gives me positive.
There's 2 tan x's there. So I've got tan squared x and then I've got tan delta x.
This is all over 1 minus tan x tan delta x and remember,
this entire thing is now over this delta x as well.
So, this is the answer to this part here and then we also have delta x at the bottom.
Now the good thing about this is, is that if you have 2 things dividing,
so, I have a denominator and another denominator there in fact multiplying together,
so you can actually join them. This actually joins together with that value there at the bottom.
So, I can just rewrite this because it's just delta x at the bottom and it makes life a little bit easier.
I can just write delta x here and that's my entire fraction now so, I've got that all together.
So there's nothing that I've missed out on. Okay, good things are happening.
You can see here you've got a tan x and a minus tan x, so we've done something.
We've managed to simplify a little bit. If I just rewrite what I have,
I have tan delta x left, plus tan squared x tan delta x all over delta x 1 minus tan x tan delta x.
So, this is what I've left with. The good thing to do perhaps is to factorize it.
So maybe we can take a factor, a common factor of tan delta x out here.
So we've got that in common and we can rewrite this as tan delta x.
This whole equation is over delta x so I'm going to take that delta x out as well.
Now, leaving me with 1 plus tan squared x over 1 minus tan x tan delta x.
You don't have to do this step but you'll see that it makes things a little bit easier
as you try and calculate the limit. Right, a thing that we need to make sure that we have is this,
that the limit of delta x tends to zero. So, the limit of delta x also needs to tend to zero here,
and now we can apply that to each part separately because they're both products
so we can apply this limit to both products here. If you take tan of delta x over delta x,
by the same sort of argument as the sine, this also equals to 1 and this second part,
when you change this, this doesn't get affected by delta x.
So we have 1 plus tan squared x and you will notice that as delta x tends to zero,
this entire term becomes zero because of the delta x. So you have, 1 minus zero at the bottom.
We're eventually left with 1 plus tan squared of x and if you look back at the identities
that we gave earlier, you know that 1 plus tan squared of x is the same as sec squared of x
which is the derivative or the answer to tan x when you're differentiating it.
So, fairly complicated proof algebraically and using identities
but it does cover our large chunk of mathematics because we're looking at trigonometry,
we're looking at identities, we're looking at differentiation,
we are also looking at differentiation from first principles.
So there's a lot of skill and a lot of calculus in doing these proofs.
So well done if you've managed to do them by yourself.
If not, try it out so you can actually derive them.