WEBVTT

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In previous tutorials, you've
differentiated from first

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

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

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and log to the base e of x.

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

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

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

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derivatives and add those to
our table.

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So let's start with a function.

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f of x.

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And add derivative.

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

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

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

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x, the derivative of x is 1.

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2x, the derivative is 2.

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If we have x to the power n.

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

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where n is a real number.

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sine x, the derivative is cos x.

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cos x, the derivative is -sine x.

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

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And log to the base e of x.

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Where the derivative is 1 over x.

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

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back to the table in just a minute.

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

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we have 2 sine x.

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Well, this is just the same as

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taking sine x and taking two of them.

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So when we find the derivative,
we actually find the

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derivative of sine x and take
two of those.

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Well, the derivative of sine x
is cos x.

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

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

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f dashed of x is -5 cos x.

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And this in fact is the case for

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any constant multiplied by the function.

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

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

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And this is called the
constant multiplier rule.

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So if we have...

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C times our function of x.

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

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derivative with respect to x.

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

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C times our f dashed of x.

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Now let's prove this from
first principles.

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

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

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Now, our definition of our derivative

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

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

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

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

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So we substitute that in.

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

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All divided by delta x.

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

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have the limit as delta x
approaches zero of

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

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divided by delta x.

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And because our constant is
nothing whatsoever to do with

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

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

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

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divided by delta x.

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

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so it's actually our f dashed of x.

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So our g dashed of x, our derivative,

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is equal to C times our f dashed of x.

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

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we have two functions that
are added together.

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

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And we want to differentiate
them with respect to x.

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

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

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separately and add them together.

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

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Similarly, what happens if we
want two functions that are

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subtracted and we want to
differentiate those?

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

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It's very straightforward.

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Because what we've got here.

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

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function multiplied by minus one.

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And there we are using the

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

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derivative of d of x of f of x plus

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minus one times the derivative of g of x.

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So that's our df dx plus

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our minus one times our 
derivative there which is

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plus minus one times dg dx.

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

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Let's just return to our table.

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And we can write those in.

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

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is c times df dx.

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

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Our derivative is df dx plus dg dx.

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

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Then the derivative
is df dx minus dg dx.

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Now let's have a look at
an example.

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2 x cubed minus 6 cos x

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

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And that equals the derivative of

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2 x cubed with respect to x, minus

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the derivative of 6 cos x 
with respect to x.

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

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

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

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and again here using the constant
multiplier rule, we can

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take the six outside.

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

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So we have twice, now the
derivative of x cubed

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with respect to x,

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is 3 x squared minus, 6 times

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

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

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positive times 6 sine x.

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Let's look now at extending
the table further.

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

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And our derivative.

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df dx or f dash of x.

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

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any constant number

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and the cos of mx.

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

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And then log to the
base e of mx.

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

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

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u is equal to mx.

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So therefore our y is
equal to sine u.

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

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So du by dx is equal to m.

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

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u. So dy by du equals

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the derivative of sine u,
which is cos u.

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

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dy by dx is equal to

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equal to dy by du multiplied by
du by dx.

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Now this is called differentiating
a function of a function

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

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

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

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

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

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Now we usually write the
constant first, so

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it'll be m times the cos of u.

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

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

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

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

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

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we've multiplied by m.

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

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The derivative of sine mx
is m cos mx.

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

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

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

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So our y equals x to the 
power of mx.

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Again, we're going to do a substitution

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and let u equal mx.

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

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

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

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And we get e to the u.

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

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

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du by dx is m.

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

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

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

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

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Let's put that back in our table.

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So e to the power of mx.
The derivative is

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m e to the power of mx.

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

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So y equals log to the base e
of mx.

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

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

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Our du by dx equals m.

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

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So our dy by dx equals

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dy by du multiplied by
du by dx.

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

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

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conveniently here, our m's cancel out

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

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So let's put that now in our

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

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

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

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logarithm of a negative number.

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

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

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

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by using the laws of logarithms.

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

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

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And then differentiated them.

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

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

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

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is one over X.

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

00:18:22.165 --> 00:18:23.593
to one over X.

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

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the base E of AX plus B.

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

00:18:48.930 --> 00:18:52.269
So Y equals log to the base.

00:18:52.950 --> 00:18:55.536
A of a X plus B.

00:18:57.440 --> 00:19:00.261
Again, we're going to
use the substitution.

00:19:00.261 --> 00:19:02.679
You equals a X plus B.

00:19:04.120 --> 00:19:06.937
So why? Why is it cool to log to

00:19:06.937 --> 00:19:09.310
the base E? Of you.

00:19:12.280 --> 00:19:16.272
We differentiate you
with respect to X. Do

00:19:16.272 --> 00:19:18.767
you buy DX equals a?

00:19:20.430 --> 00:19:24.800
At the Y fi do you
equals one over you?

00:19:26.490 --> 00:19:33.310
Again, I DYIDX is equal
to DY by du multiplied

00:19:33.310 --> 00:19:36.720
by to you by DMX.

00:19:38.070 --> 00:19:40.630
The Wi-Fi do you
is one over you.

00:19:42.250 --> 00:19:46.669
Multiplied by do you buy DX,
which is a?

00:19:48.030 --> 00:19:55.386
So we have a divided by
IU, which is a X plus

00:19:55.386 --> 00:19:55.999
B.

00:19:57.670 --> 00:19:59.760
So do why by DX.

00:20:00.180 --> 00:20:03.789
Sequel to a divided by a
X plus B.

00:20:04.810 --> 00:20:07.434
So finally, let's add that
to our table.

00:20:10.030 --> 00:20:16.232
The derivative of locked the
base E of X Plus B is a divided

00:20:16.232 --> 00:20:18.004
by 8X Plus B.