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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
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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.
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So Y equals log to the base.
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A of a X plus B.
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Again, we're going to
use the substitution.
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You equals a X plus B.
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So why? Why is it cool to log to
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the base E? Of you.
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We differentiate you
with respect to X. Do
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you buy DX equals a?
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At the Y fi do you
equals one over you?
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Again, I DYIDX is equal
to DY by du multiplied
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by to you by DMX.
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The Wi-Fi do you
is one over you.
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Multiplied by do you buy DX,
which is a?
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So we have a divided by
IU, which is a X plus
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B.
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So do why by DX.
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Sequel to a divided by a
X plus B.
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So finally, let's add that
to our table.
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The derivative of locked the
base E of X Plus B is a divided
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by 8X Plus B.