www.mathcentre.ac.uk/.../8.3%20Use%20table%20derivs_8functions.mp4

In previous tutorials, you've
differentiated from first

principles functions such as
c, x, 2x, x to the power n,

where n is any power, sine x,
cosine x, e to the power x

and log to the base e of x.

Now what we're going to do in
this tutorial is to construct

a table of these standard
derivatives, and when we've done

that, we'll look at some
variations of those standard

derivatives and add those to
our table.

So let's start with a function.

f of x.

And add derivative.

Which may be referred to as,
df by dx or f dashed of x.

So let's start with c, well c is a
constant. It's a straight line

and it's horizontal. So I
gradient function is 0.

x, the derivative of x is 1.

2x, the derivative is 2.

If we have x to the power n.

Then the derivative is n times x
to the power of n - 1,

where n is a real number.

sine x, the derivative is cos x.

cos x, the derivative is -sine x.

e to the power x, that's the one that
stays the same, e to the power x.

And log to the base e of x.

Where the derivative is 1 over x.

Let's look now at a variation
of some of these, we'll come

back to the table in just a minute.

OK, what if we have f of x
equals, instead of just sine x,

we have 2 sine x.

Well, this is just the same as

taking sine x and taking two of them.

So when we find the derivative,
we actually find the

derivative of sine x and take
two of those.

Well, the derivative of sine x
is cos x.

So the derivative of 2 sine x
is 2 cos x.

Similarly, if we take f of x is
-5 sine x, then our derivative

f dashed of x is -5 cos x.

And this in fact is the case for

any constant multiplied by the function.

So if we have f of x equals
C times sine x.

Then f dashed of x is going to
be C times cos x.

And this is called the
constant multiplier rule.

So if we have...

C times our function of x.

And we want to differentiate it.
So we want to find the

derivative with respect to x.

It's actually just C times the
derivative df by dx or

C times our f dashed of x.

Now let's prove this from
first principles.

If we take our function of x
to be g of x and

it's equal to some constant times f of x.

Now, our definition of our derivative

is g dashed x equals the limit
as delta x approaches 0 of

g of x plus delta x, minus g of x,
all divided by delta x.

Which equals, again our limit
as delta x approaches 0.

And in this case our g of x plus delta x
is C of f of x plus delta x.

So we substitute that in.

Minus our g of x which is C f of x.

All divided by delta x.

And here I'm going to take the
C outside the bracket so we

have the limit as delta x
approaches zero of

C and then our function of x plus delta x
takeaway our function of x...

divided by delta x.

And because our constant is
nothing whatsoever to do with

our limit is delta x approaches
zero we can actually take the

constant outside of the limiting
sign, so the limit delta x approaches 0

and we have f of x plus delta x,
minus our function of x...

divided by delta x.

So this is C and this...
is our derivative

so it's actually our f dashed of x.

So our g dashed of x, our derivative,

is equal to C times our f dashed of x.

Now let's have a look at what
we're going to do when

we have two functions that
are added together.

So let's say we have
f of x, plus, g of x.

And we want to differentiate
them with respect to x.

So we want the derivative of
that function with respect to x.

Well, quite simply, what we do
is we differentiate each part

separately and add them together.

So that's the same as
df dx plus dg dx.

Similarly, what happens if we
want two functions that are

subtracted and we want to
differentiate those?

Well, again, the derivative of those
functions subtracted...

It's very straightforward.

Because what we've got here.

Is just the same as we've got
here. But with this second

function multiplied by minus one.

And there we are using the

constant multiplier rule. So
this is just the same as the

derivative of d of x of f of x plus

minus one times the derivative of g of x.

So that's our df dx plus

our minus one times our
derivative there which is

plus minus one times dg dx.

Which is just the same as df dx
minus dg dx.

Let's just return to our table.

And we can write those in.

So we have our constant multiplier
rule: c times our function of x,

is c times df dx.

And if we have f of x plus g of x.

Our derivative is df dx plus dg dx.

And if they're subtracted f of x
minus g of x.

Then the derivative
is df dx minus dg dx.

Now let's have a look at
an example.

2 x cubed minus 6 cos x

And we want to find the
derivative with respect to x.

And that equals the derivative of

2 x cubed with respect to x, minus

the derivative of 6 cos x
with respect to x.

Now what we have here is the 2,
which is,

we can use the constant multiplier rule.
So that we've got actually:

Twice the derivative of x cubed
with respect to x, minus,

and again here using the constant
multiplier rule, we can

take the six outside.

So it's six times the derivative
of cos x with respect to x.

So we have twice, now the
derivative of x cubed

with respect to x,

is 3 x squared minus, 6 times

the derivative of cos x, which is
minus sine x.

So we have two threes, 6 x squared,
minus times minus is

positive times 6 sine x.

Let's look now at extending
the table further.

So let's start with our function
f of x again.

And our derivative.

df dx or f dash of x.

And this time let's look at sine
of mx where m is

any constant number

and the cos of mx.

Then we'll have a look at e to
the power of mx.

And then log to the
base e of mx.

Let's look then at y equals sine
mx. Now here we're going to do a

substitution and instead of mx,
we're going to write

u is equal to mx.

So therefore our y is
equal to sine u.

And we're going to differentiate
u with respect to x.

So du by dx is equal to m.

And then we're going to
differentiate y with respect to

u. So dy by du equals

the derivative of sine u,
which is cos u.

Now dy by dx, and you'll
learn this in the future,

dy by dx is equal to

equal to dy by du multiplied by
du by dx.

Now this is called differentiating
a function of a function

or differentiating using the chain rule,
and it's the subject of another tutorial.

So we're not going to go
through the details of that now.

What I want you to do
is just use it as a formula.

Let's now substitute then
dy by du is equal to

cos u multiplied by du by dx,
which equals m.

Now we usually write the
constant first, so

it'll be m times the cos of u.

But we introduced the u and we
want actually dy in terms of dx

in terms of x, dy by dx.
So what we need to do is to

substitute back and instead of
writing u we want to write mx.

So we have as dy by dx equals m
times the cos of mx.

So in effect, what's happened
when we found the derivative is

we've multiplied by m.

So let's go back and
put that in our table.

The derivative of sine mx
is m cos mx.

And in a similar way, the
derivative of cos mx will be

minus m sine mx again, just like
multiplying by the m.

Let's look now at e
to the power of mx.

So our y equals x to the
power of mx.

Again, we're going to do a substitution

and let u equal mx.

So that our y is equal to e
to the power u.

Again, we're going to
differentiate u with respect to x.

I'm gonna get m and will differentiate
y with respect to u.

And we get e to the u.

Our dy by dx is equal to
dy by du multiplied by du by dx.

And if we substitute in, dy by du is
e to the u.

du by dx is m.

Again, let's write our
constant terms first, so

we get m e to the power u.
Now we need to

substitute back for our u so
we get m e to the power of mx.

So again, our derivative we've
ended up multiplying by m.

Let's put that back in our table.

So e to the power of mx.
The derivative is

m e to the power of mx.

Let's look now at log to the
base e of mx.

So y equals log to the base e
of mx.

Same process again.
Let's let u equal mx.

So our y equals log to
the base e of u.

Our du by dx equals m.

dy by du, the differential of
log to the base e of u is 1 over u.

So our dy by dx equals

dy by du multiplied by
du by dx.

Our dy by du is 1 over u.
Our du by dx is m.

So we have m divided by u,
which is mx and very

conveniently here, our m's cancel out

and we end up with 1 over x.
So dy by dx is equal to 1 over x.

So let's put that now in our

table. So log to the base E of
MX is our function, so our

derivative is one over X and of
course this is only the case

where M is greater than zero.
Since we can't take the

logarithm of a negative number.

Let's just go back to the
calculation that we've just

done, because there is another
way that we could have looked

at. Differentiating Y equals log
to the base E of MX and that was

by using the laws of logarithms.

What we could have done was
said that Y equals log to

the base E of M plus log to
the base E of X.

And then differentiated them.

So I DY by DX is equal to
what log to the base E of M.

Is just a constant. So when we
differentiate that we get 0

plus. Log to the base E of X
where we know the derivative

is one over X.

So we end up with just the same
as before divided by DX is equal

to one over X.

Not one final one to
have a look at is log to

the base E of AX plus B.

So let's just have
a look at that one.

So Y equals log to the base.

A of a X plus B.

Again, we're going to
use the substitution.

You equals a X plus B.

So why? Why is it cool to log to

the base E? Of you.

We differentiate you
with respect to X. Do

you buy DX equals a?

At the Y fi do you
equals one over you?

Again, I DYIDX is equal
to DY by du multiplied

by to you by DMX.

The Wi-Fi do you
is one over you.

Multiplied by do you buy DX,
which is a?

So we have a divided by
IU, which is a X plus

B.

So do why by DX.

Sequel to a divided by a
X plus B.

So finally, let's add that
to our table.

The derivative of locked the
base E of X Plus B is a divided

by 8X Plus B.