So the next question that we're going to answer is what does
differentiation actually do.
Why are we differentiating functions and why are we using
these complicated and easy methods to find derivatives.
Differentiation helps you to find the gradient of a curve
at a particular point.
You may think that's a bit strange because the gradient
of a curve changes all the time.
But you can find the gradient of a particular point on the curve
just by substituting your x and y values into that point.
It really is quite amazing how that works.
That when you differentiate, you get a general gradient
of the curve.
And then you can look at any specific point by just
substituting numbers into it.
We'll try this out with some examples shortly.
First of all we'll look at practicing A,
So we can do this really quickly now because I won't make you
do this using first principles.
We're just going to make sure that we can differentiate
and differentiate correctly.
And then we'll move on to substituting numbers
in finding the gradient at more specific points.
And eventually we'll move on to finding gradients
of tangents and lines and even equations of tangents of lines
at those specific points.
So let's have a look at this question.
We will differentiate y = 3x2 + 5x - 1
and we'll do it the faster way.
So remember what we said that dy by dx.
You bring the power to the front.
So we'll have 2.
3x, 3 stays as it is.
Decrease the power of x by 1.
If I write it out as the whole thing.
You still bring the power of x down.
So the power of x now is 1.
And then you have 5x to the 1 - 1
which just becomes x to the 0.
And then the constant in the end will disappear
because it has a power of x to the 0.
This than gives you an answer of 6x + 5.
and that's your gradient.
So really that's now made our lives a lot easier
because we don't have to apply differentiation the first
principles, we can just do it the faster way.
Remember now that this isn't just dy by dx.
This is actually telling us something.
It's telling us the gradient of this equation.
It's telling us how steep it is and at any different,
any given point.
So say for example if this is a curve.
And the equation to that curve is y = to 3x2 + 5x - 1.
If we were to sketch this curve, you could factorize this.
So we've got 3x and x.
And then you'll have 1 and 1.
If you try this out, you will see that no matter what signs
that you put in here and you won't quite be able to factorize it.
So they are other methods if you recall.
So previously if you remember that you would have done
the quadratic formula which states
the x = -b+-squared(b-4ac)/2a.
Your a is the number next to the x squared.
So this is your a.
Where b is the number next to the x
and c is the number by itself.
So if ever you are not able to factorize to find points of
intersections of a curve, remember there's always backup
plans that we could use this equation to find our solutions.
So X is equal to -5, +- square root of 5 squared, -4 X 3, multiplied by -1.
This is all over 2 x 3, x is equals to -5 as before,
and then in the root 5 squared gives us 25 and -4 x 3, -1, gives us +12.
And this is still all over 6.
So we end up with -5 +- root of 37 all over 6.
There's no obvious simplication factors here, so we can just leave it as it is.
Therefore, we have two answers, one positive and one negative, due to the +- before the root.
So if I sketch this, just giving over a rough idea.
So we got a positive and a negative answer.
And we've got a quadratic.
What we just found for ourselves is that the gradient
of this curve at any given point, so here or here or here,
And all you have to do is to find the actual gradient at that
point is to substitute this x value into this equation.
So you could find the gradient at this point.
If you know this x value, you're just substituting here,
or you could find the gradient at this point.
You know the x value you just substitute in here to find your
specific gradient at that point.
Let's look at our second example now.
Now this is the only way that they can really make our lives
difficult, is by giving us differential or equations to
differentiate which are little bit more complicated.
So you have square roots or you have x values at the bottom
And how do we tweak it and how do we make it more
differentiation friendly for ourselves.
So let's look at this question and see if we can
make it a little bit easier.
And just apply the straight forward concepts
So I'm just going to write this question again
so that we can move things around.
So we've got y = 4 over x cubed,
plus 2x to the power of 5 minus root x.
Now the first thing that we need to do in order to make our lives
little bit easier, is we don't want this as denominator.
So we need to bring this up.
And secondly we don't want things like these.
So we don't want roots.
We want powers.
Because powers we can deal with and we know what to do
with the power of something.
We can bring it down and we can decrease that by 1.
So before I differentiate I'm going to rewrite this.
As you know by the rule of indices
or the rule of denominators being at the bottom,
you are allowed to bring this up.
So you can bring this value up, as long as you change
the sign of the power.
So we can rewrite this as 4x to the minus 3.
So this is one of the rules of the indices.
So we've got 2x to the power 5 which doesn't bother us.
And then we've got x to the power of a half.
So remember now I haven't differentiated it.
I've just rewritten it in order to make it easier for myself.
And this is perhaps the only way that these differentiation
questions can become a little bite mathematically challenging.
But all you have to do is just look at it,
simplify a little bit for yourselves.
Make it more differentiation friendly and then we just apply
the same rule over and over again no matter
what they are asking.
So here if I dy by dx and then
I'm going to bring the power down so minus 3.
I've got 4x to the minus 3 of minus 1.
So decrease the power by 1.
You can bring 5 down.
The 2 is a constant which stays as it is.
And then you take 1 away from it.
And now we bring the half down and remember it doesn't matter
what the power is.
No matter how complicated that is.
You just apply the same rule to it.
And then I've got x to the half minus 1.
And then I've basically done the difficult thing.
All I have to do is tidy it up and that will give me
my final answer.
So this gives me -12 when I multiply these two numbers
I have x to the minus 4.
5 times 2 gives me 10.
x to the 4.
And then I have minus the half x to the minus the half.
And if you ever struggle with working something like this out,
you can always just do it on the side because you are not
quite using a calculator a half minus 1.
You want 2 to be your common denominator.
So you want both of these numbers to be above 2.
This changes, this doesn't change any.
This stays as it is.
And in order to increase this whole fraction by 2,
you multiply the bottom, with the top by 2 as well.
So that gives you minus 2 to give you minus 1 at the top
and 2 at the bottom.
So that's where it comes from.
So if fractions ever become difficult and you're not using
a calculator, there's no shame in working out on the side
and then putting it back into your answer.
Final few things that we could do so,
we can make this x to the minus 4 go down.
So we can make it positive, so that becomes x to the 4.
Plus 10x to the 4.
And then we can bring this negative x to the minus
the half down.
So that becomes joins the 2 at the bottom.
So that becomes the half.
And if you rewrite this as a square root,
we can also do that.
And all you can do now is just change this last setup
here if you wanted too.
So you could write this as square root of x.
Just the way that they did in the question.
So we have 12 over x to the 4 plus 10x to the 4
and then -1 over 2.
And instead of x to the half, you can write as root x just to
show that you understand where it comes from.
So once again you are using a rule of indices to convert it
back and make it look tidier.
So the gradient of this complicated function here
is given by this.
I'm not going to try and sketch this out.
But remember that you can find the gradient at any particular
point on this curve just by making substitutions for these
x values here.
So if for example you are looking at the gradient at x
equals to 2.
All you have to do is substitute 2 instead of the x's
and you will get your gradient numerical value,
be it positive or negative.
It can tell you a lot about what the curve is doing.