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.