Differentiation — 1/11
Differentiation
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Differentiation — 2/11
Now back to the notion of differentiation. We then define
the differential of y(x) as,
Contents
1
Introduction
2
2
Power rule
2
3
Table of derivatives
4
4
Curve sketching
4
5
Chain rule
6
6
Product Rule
8
7
Quotient rule
8
Examples
9
11
1. Introduction
One of the most important aspects about differentiation is to
grasp the initial ideas and then build upon them. The skills and
techniques learnt are not difficult, but once you can spot which
technique to use in a question then the problem becomes much
simpler to solve.
Lets begin with some of the fundamental ideas of differentiation. We start off with some basic graph sketching, so
we define some equation by,
y = x+3
Then we say that this variable y depends on some other variable x. So we say that y is some function depending on the
variable x, and thus denote it as y = y(x). That is, for every
point x we can find a corresponding y(x) value. Therefore,
we must be able to construct a graph or make a plot of such
a function. The variable x is also known as an input value;
something that we insert into the equation. Then for this value
of x, we find that the corresponding output value is given as
y(x).
We then take a few points as an example,
x = 2 → y(2) = 2 + 3 = 5
and then sketch this function with the sets of points we have
determined, (0, 3), (1, 4), (2, 5).
4
2
2
-4
this is also known as a linearity property. Secondly, if we have
a different function this time defined as y(x) = cu(x) where c
is just a constant term and u(x) is some function of x. Then,
dy(x) d(cu(x))
du(x)
=
=c
dx
dx
dx
where all that we have done has been to take out the constant
term outside.
2. Power rule
The first main technique is simple and probably the most
important. It is known/refereed to as the power rule. If we
have a function defined the following way,
(1)
dy(x)
= y0 (x) = nxn−1
dx
(2)
where n is just a number, it could be 3, 4, 1/2, 100 etc. but
not a variable of x. So in other words, n 6= n(x). This would
be slightly more complicated. So for example, if we chose
n = 2 then we would simply replace it in equation (1) and use
the rule given by (2),
yHxL
-2
d
dy(x)
=
y1 (x) + y2 (x)
dx
dx
dy1 (x) dy2 (x)
=
+
dx
dx
Then the differential of this function with respect to the variable x is given by,
x = 1 → y(1) = 1 + 3 = 4
-2
The left hand side reading; the differential of y(x) with respect
to the variable x. The right hand side being exactly the same
only written in a shorthand notation.
Before we jump into the practical side of things, we note
a few common and useful properties of differentiation. These
might not make much sense now but after some practice they
may become clearer and worth a revisit. Sometimes we even
take these properties for granted if we have learnt differentiation from another course.
If we have a function y(x) = y1 (x) + y2 (x), then
y(x) = xn
x = 0 → y(0) = 0 + 3 = 3
-4
dy(x)
= y0 (x)
dx
4
x
y(x) = x2
dy(x)
= y0 (x) = 2x(2−1) = 2x1 = 2x
dx
So we have essentially multiplied down by 2 and subtracted 1
from the power. The same works if we have extra terms added
Differentiation — 3/11
onto equation (1). This is precisely what the linearity property
allows us to do. So defining,
y(x) = x2 + x3
dy(x)
= 2x + 3x2
dx
where we have used the exact same rule (2) twice this time
for both n = 2 and n = 3. Now, looking at something a bit
trickier, y(x) = x. We do exactly the same as before,
y(x) = x1
dy(x)
= 1.x1−1 = (1)(x0 ) = (1)(1) = 1
dx
If you can remember back to the topic of indices that anything
to the power of 0 is just reduced to 1 (so x0 = 1).
But what happens when we have a constant term (i.e. just
a number) to differentiate? Again, we apply what we already
know given by (2),
y(x) = 7x + 3
y(x) = 7x1 + 3x0
dy(x)
= 7.(1)x1−1 + 3(0)x0−1 = 7x0 + 3(0)x−1 = 7
dx
where we used a little trick of 3 = 3x0 as we recall from
indices. Also, the second term is multiplied by zero towards
the end. So regardless of the 3 and x−1 , they are just multiplied
by 0 which means that the whole term vanishes.
So far we have just discussed the possibility of whole numbers for n, but what if we were to consider fractions, negative
numbers or even negative fractions! Thankfully all of these
values for n work exactly the same way as before but just a
little harder. Below are a couple of examples but it will all
come down to practice makes perfect.
1. Negative numbers
We look at the easiest case first of n = −1
y(x) = x−1
dy(x)
1
= (−1)x(−1−1) = −x−2 = − 2
dx
x
Let’s look at a second example of the following form,
1
x3
First we must rearrange this fraction, we can’t just differentiate
it as it is, e.g.
y(x) =
dy(x)
1
6= 2
dx
3x
This is incorrect and we will see why that is so. We must
rewrite y(x) in the following way first,
y(x) =
1
= x−3
x3
1
dy(x)
3
= −3x−4 = − 4 6= 2
dx
x
3x
2. Fractions
Next we look at some powers that are now fractions such as,
y(x) = 2x3/2
3
dy(x)
=2
x(3/2−1) = 3x(3/2−2/2) = 3x1/2
dx
2
again, all this was accomplished using the same rule as we
mentioned at the beginning of this section given in equation
(2). The only slip-up that may occur in these examples is a
mistake in working with fractions. Here is another example
that may look hard but can be taken step by step easily,
y(x) = x17/10 + 40x1/3
1
dy(x) 17 (17/10−1)
=
x
+ 40
x(1/3−1)
dx
10
3
dy(x) 17 (17/10−10/10) 40 (1/3−3/3)
= x
+ x
dx
10
3
dy(x) 17 7/10 40 −2/3
= x
+ x
dx
10
3
dy(x) 17x7/10
40
=
+ 2/3
dx
10
3x
2. Negative fractions
The following is just simply a combination of both points 1.
and 2. In order to see this work we just use an example for an
explanation. We take,
y(x) =x−4/3 − 10x−1/2
4
−1 dy(x)
=−
x(−4/3−1) − 10
x−1/2−1
dx
3
2
4
10 dy(x)
=−
x(−4/3−3/3) +
x−1/2−2/2
dx
3
2
4
x−7/3 + 5x−3/2
=−
3
4
5
= − 7/3 + 3/2
3x
x
Differentiation — 4/11
3. Table of derivatives
4. Curve sketching
Aside from the main definition of differentiation given by (2)
and the rules that will shortly follow, we need to first define
some more complicated differentials of functions. These will
be required when we discuss the remainder of the rules later
on when tackling more difficult examples.
Most of the basic and commonly used functions have been
differentiated in Table 1. We will use these on a regular basis
so be sure to know these. These are widely accepted to use,
there is however a strict definition to determine these. We will
not look at this definition here as it is beyond the scope of this
booklet.
Depending of your background/situation, generally we all
know how to sketch a graph if we are given a basic function.
Don’t worry if you can’t however, as we will go through some
simple examples first. Generally for a linear equation (highest
order of x is to the power 1, a quadratic equation has highest
power of 2 etc.) we can write
y(x)
dy(x)/dx
sin(x)
cos(x)
cos(x)
− sin(x)
tan(x)
sec2 (x)
ln(x)
(x)0 /x
x
e
ex
ax
0
ax
e
(ax) e = aeax
In the table when we write something in the form of (ax)0 , this
means that we differentiate (ax) with respect to x. In other
words,
d(ax)
=a
dx
y = mx + c
(3)
where m is the gradient of the slope (how steep the line is)
and c is the intercept (where it crosses/cuts the y-axis). As a
basic example we consider,
y(x) = x
where here we have set m = 1 and c = 0, so that the line will
cross through the point at x = 0 (the intercept point). Or it can
also be constructed by simply selecting some points such as,
y(0) = 0 → (0, 0)
y(1) = 1 → (1, 1)
Some of us may even know two other important graphs such
as y = ±x2 given in Figure 1. and Figure 2.
yHxL
If you can remember, this is exactly one of the properties we
noted at the start where we could take a constant term outside
the differential. Or if you prefer the usual notation which we
have adopted, let y(x) = ax and then,
10
8
6
dy(x)
=a
dx
4
2
both are equally as valid to write.
-4
2
-2
4
x
4
x
-2
-4
Figure 1. Graph of y(x) = +x2
yHxL
4
2
-4
2
-2
-2
-4
-6
-8
-10
Figure 2. Graph of y(x) = −x2
Differentiation — 5/11
But what if some of the functions weren’t as easy to
sketch? Would you be able to sketch,
y(x) = x3 − 3x2 − 9x + 2
(4)
This is where differentiation comes in handy, we are able to
use its tools to draw more complicated equations like the one
above.
In order to sketch this graph we first need a couple of basic
points. These help us to accurately draw the graph whenever
we determine its nature (i.e the shape of the curve). Generally
a good starting point is to look at 2 cases, when x = 0 and
when y(x) = 0. In other words, when the curve cuts the y-axis
and x-axis respectively. So when x = 0,
y(0) = 03 − 3(02 ) − 9(0) + 2 = 2
2
x − 3x − 9x + 2 = 0
The above as you can imagine is not easy to solve, so we just
simply attempt different x-values.
x =0 → y(0) = 03 − 3(02 ) − 9(0) + 2 = 2 6= 0
x =1 → y(1) = 13 − 3(12 ) − 9(1) + 2 = −9 6= 0
x = − 1 → y(−1) = (−1)3 − 3(−1)2 − 9(−1) + 2 = 7 6= 0
x =2 → y(2) = 23 − 3(22 ) − 9(2) + 2 = −20 6= 0
3
2
x = − 2 → y(−2) = (−2) − 3(−2) − 9(−2) + 2 = 0 = 0
Generally if the first 5 or so terms don’t work out then I would
strongly advise to stop and just carry on with the calculation.
It won’t be the end of the world! This only helps us gather
some extra information.
The next step is to perform the differentiation of y(x) given
by (4) with respect to x,
dy(x)
= 3x2 − 6x − 9
dx
dy(x)
=3x2 − 6x − 9 = 0
dx
x2 − 2x − 3 = 0
(x − 3)(x + 1) = 0
x = 3 & x = −1
These two values for x are simply values that are points lying
on the curve of y(x) in equation (4). The only difference is
that they are unique and very useful. So these two points
should provide us with corresponding y(x) values when we
substitute them into (4),
y(3) =33 − 3(32 ) − 9(3) + 2
Coordinates are represented in the form of (x, y), so (0, 2) is
our first set of points. However, generally it is not as easy
to solve for y(x) = 0. This turns into a guessing game for
x-values until we get the right value (although sometimes we
might not even obtain a set of points!). So, imposing y(x) = 0
we are then required to solve,
3
is equal to 0 for a value of x. A gradient of zero graphically
represents a horizontal line, it slopes neither up nor down.
That means we set dy(x)/dx given by (5) equal to 0,
(5)
The reason this is important is because dy(x)/dx is essentially
equal to m in equation (3), denoting the gradient. However,
what dy(x)/dx describes is very useful. It tells us what the
gradient of the curve y(x) is at any point x. So as an example
if x = 1 then the gradient of the curve would take the value,
dy(x)
= 3 − 6 − 9 = −12
dx
In other words, the gradient is always changing between different x-values. We can exploit this knowledge to find where the
curve of y(x) turns. In other words, the gradient of the curve
=27 − 27 − 27 + 2
= − 25
y(−1) =(−1)3 − 3(−1)2 − 9(−1) + 2
=−1−3+9+2
=7
So (3, −25) and (−1, 7) are our two turning points. The final
step before we begin to sketch the graph of y(x) is to determine the nature of these turning points. In order to do so
we must compute the second derivative. This is defined the
following way,
First derivative
As we would imagine is,
dy(x)
dx
Second derivative
Basically taking the derivative of the first derivative,
d dy(x) d 2 y(x)
=
dx dx
dx2
So all we need to do is to just different dy(x)/dx given by (5)
with respect to x once more,
d 2 y(x) d dy(x) =
dx2
dx dx
d
= (3x2 − 6x − 9)
dx
=6x − 6
The definition is then given as:
The turning point is a minimum when,
d 2 y(x)
>0
dx2
and the turning point is a maximum when
d 2 y(x)
<0
dx2
Differentiation — 6/11
We recall here that our two turning points have been determined as (3, −25) and (−1, 7), so all we do is just substitute
2
our values of x = 3 and x = −1 into d dxy(x)
2 and see which value
gives rise to a positive value and which gives rise to a negative
value.
d 2 y(x)
= 18 − 6 = 12 > 0
dx2
So (3, −25) is a minimum.
2. For x = 3
dx2
dy(x)
= 3u2
du
(6)
But the aim of this problem is to determine what y’(x) is, and
clearly,
1. For x = 3
d 2 y(x)
which we recognise as being easily differentiable. We can’t
however differentiate this respect to x but with respect to the
new variable u. In other words,
dy(x) dy(x)
6=
dx
du
However, it is true that,
dy(x) dy(x) du
=
.
dx
du dx
= −6 − 6 = −12 < 0
So (−1, 7) is a maximum.
Finally we have everything we need. Now just to summarise everything that we have calculated so far.
(0, 2)
(7)
which is precisely the definition of the chain rule if we make
one substitution of u. So far we have worked out the first term
on the right hand side of equation (7). All that remains now is
to determine du/dx. We already know that we performed the
following substitution at the start:
u = 2x + 1
(−2, 0)
(3, −25) min. tp.
(−1, 7) max. tp.
In Figure 3. The two blue horizontal lines represent the minimum and maximum turning points. If you look closely as well
on the left hand diagram there are 4 blue dots representing
the 4 points we have determined. All we do is simply join up
these dots given by the dashed line on the right hand image.
A little bit of common sense comes into play when it comes
to graph sketching.
5. Chain rule
We follow up by looking at the first proper rule of differentiation which is possibly one of the easier/easiest of rules,
but one of the toughest to master. We shall hopefully see the
complexity of the rule towards the end of this section and
further in the worked examples.
The chain rule is especially helpful when we want to
differentiate a function that is within another function. For
example,
3
y(x) = (2x + 1)
We could of course multiply this bracket out and differentiate
as normal. That method however is very lengthy and time
consuming. The method involved to solve this is to make
some kind of intuitive or smart substitution.
Let,
u = 2x + 1
Then y(x) becomes,
y(x) = u3
So we simply calculate,
du
=2
dx
(8)
Now that we have found what we are looking for, we can
simply multiply both (8) and (6) together and use the chain
rule given by (7),
dy(x)
= (2)(3u2 ) = 6u2
dx
Then to obtain everything in terms of x we simply substitute
our value of u = 2x + 1 back into the above,
dy(x)
= 6(2x + 1)2
dx
which is the final answer. Let us go through how to solve
something similar without the wordy explanation to get a feel
for the method.
Example
y(x) = (3x2 + 4x1/2 + 2)4
Let u =, then,
y(x) = u4
we know from the chain rule (7) that,
dy(x) dy(x) du
=
.
dx
du dx
we then determine both these differentials separately,
1.
dy(x)
= 4u3
du
Differentiation — 7/11
yHxL
-4
yHxL
20
20
10
10
2
-2
4
6
x
-4
2
-2
-10
-10
-20
-20
-30
-30
4
6
x
Figure 3. Graph of y(x) = x3 − 3x2 − 9x + 2
2.
in previous cases we can differentiate u straight away, but now
this time is a little trickier. This is why we must introduce a
second substitution variable. So, let v = cos(x). Then,
du
2
= 3(2)x1 + 4(1/2)x−1/2 = 6x + 1/2
dx
x
u = ln(v)
then multiplying both of these results together we get,
dy(x)
2 = (4u3 ) 6x + 1/2
dx
x
In accordance with (9), the following quantities are required:
1.
and substituting u back into the above,
dy(x)
2 =4(3x2 + 4x1/2 + 2)3 6x + 1/2
dx
x
1 =8(3x2 + 4x1/2 + 2)3 3x + 1/2
x
dy(x)
= 2u
du
2.
which is very neat. I would strongly discourage expanding the
bracket out 4 times and differentiating that way. Plus, it would
be much more difficult to obtain the solution in its simplest
form given above.
The chain rule also works for the more difficult functions
found back in section 3 when we listed our table of derivatives.
We will now look at a specific example.
y(x) = {ln[cos(x)]}2
At first glance this seems to be a nasty function to differentiate
as we will soon encounter that we must apply the chain rule
not once, but twice! So we will need to introduce a second
substitution variable, say, v. Then we can extend our definition
of the chain rule in (7) to the following equation,
(9)
This can be expanded for as many substitution variables required to differentiate an equation, hopefully we will never
have to go as far as two substitutions like this particular case.
First thing is first, Let u = ln[cos(x)]. Then,
y(x) = u2
3.
dv
= − sin(x)
dx
Then we multiply these three differentials together to obtain,
dy(x) 2u =
(− sin(x))
dx
v
Example
dy(x) dy(x) du dv
=
. .
dx
du dv dx
1
du v0
= =
dv
v
v
and substituting all these quantities back for u = ln[cos(x)]
and v = cos(x) so everything is in terms of x,
dy(x) 2 ln[cos(x)]
=
(− sin(x))
dx
cos(x)
dy(x)
sin(x)
= −2 ln[cos(x)]
dx
cos(x)
dy(x)
= −2 tan(x) ln[cos(x)]
dx
Differentiation — 8/11
6. Product Rule
Now that we have covered the basics, which will be used
throughout, we can adopt new techniques to solve even tougher
problems. The next step to take is to learn the product rule.
But what exactly is the product rule?
Basically, it is what it sounds like. We have a product of
two functions that both depend on x. So we begin by defining
two new functions,
u(x) = x
v(x) = x
n
m
0
→ u (x) = nx
0
which is pretty messy and ugly either way! Now, the reason
that both ways work is because these functions are just simple
polynomials in x and that may not always be the case. We will
see now that we may need to deal with trigonometric terms,
logarithmic terms, exponentials etc which have been defined
in our table of derivatives. Just for satisfaction and ease of
mind we define the following for our two functions u and v,
u(x) = x4 → u0 (x) = 4x3
n−1
→ v (x) = mx
v(x) = 5x2 → v0 (x) = 5x
m−1
y(x) = u(x)v(x) = 5x4 x2
Then finally, we call,
Then,
y(x) = u(x)v(x) = xn xm
dy(x)
= v(x)u0 (x) + u(x)v0 (x)
dx
dy(x)
= 5x2 (4x3 ) + x4 (5(2)x1 )
dx
dy(x)
= 20x5 + 10x5 = 30x5
dx
where again, m is just some number like that of n. Now, we
cannot simply differentiate these one at a time like we did
earlier for two random functions, e.g.
dy(x)
6= nxn−1 mxm−1
dx
like one would expect. No, this time we use a special procedure called the product rule which is defined as the following:
If we have a function,
or, the easier (usual method),
y(x) = (5x2 )(x4 ) = 5x6
y(x) = u(x)v(x)
then y0 (x), the differential of y with respect to x is given by
the following,
dy(x)
du(x)
dv(x)
= v(x)
+ u(x)
dx
dx
dx
y0 (x) = v(x)u0 (x) + u(x)v0 (x)
dy(x)
= 5(6)x5 = 30x5
dx
both yielding the same result. Now we consider an example
that will definitely require the use of the product rule to solve.
(10)
Applying this rule to y(x) = xn xm , we find that,
dy(x)
= v(x)u0 (x) + u(x)v0 (x)
dx
dy(x)
= xm nxn−1 + nn mxm−1
dx
Or, if you managed to spot this earlier well done, but we could
have just as easily added the two indices in y(x). The point
of this exercise is that sometimes we can’t always do this and
must revert to the product rule.
y(x) = xn+m
dy(x)
= (n + m)xn+m−1
dx
Although this might not seem like the same result we have
achieved, we simply rearrange a little to find that they are
indeed the same.
dy(x)
= (n + m)xn+m−1
dx
dy(x)
= (n + m)xn xm x−1
dx
dy(x)
= nxn xm x−1 + mxn xm x−1
dx
dy(x)
= nxm xn−1 + mxn xm−1
dx
Example
Let
y(x) = x sin(x)
This is where we have to use our initiative and define u(x) and
v(x) ourselves. So we let,
u(x) = x → u0 (x) = 1
v(x) = sin(x) → v0 (x) = cos(x)
So we simply apply the product rule defined by equation (10)
and substitute our values in that have been determined above,
dy(x)
= v(x)u0 (x) + u(x)v0 (x)
dx
dy(x)
= (sin(x))(1) + (x)(cos(x))
dx
dy(x)
= sin(x) + x cos(x)
dx
Differentiation — 9/11
Then,
7. Quotient rule
This final rule is not much more different and difficult than that
of the product rule discussed in the previous section. We will
jump into the definition straight away: If we have a function
y(x),
y(x) =
u(x)
v(x)
dy(x) x2 − 2x2 − 8x
=
dx
x4
dy(x) −x2 − 8x
x(8 + x)
(8 + x)
=
=−
=−
dx
x4
x4
x3
Below we will give an example of the quotient rule along with
the chain rule discussed in Section 5.
then the differential of y(x) is given as,
dy(x) v(x)u0 (x) − u(x)v0 (x)
=
dx
v(x)2
dy(x) (x2 )(1) − (x + 4)(2x)
=
dx
(x2 )2
(11)
which is very similar to the product rule.
We first look at something extremely basic of which the
answer is easy to determine. The reason for this is just to
check that the definition works for a simple case.
Example
Let us define,
y(x) =
(x + 5)7
cos(x)
Then we can define our variables u(x) and v(x) as,
Example
Let,
u(x) = (x + 5)7 → u0 (x) = ...
v(x) = cos(x) → v0 (x) = − sin(x)
y(x) =
x2
x
or y(x) = x, which the differential we know is just simply,
dy(x)
=1
dx
here we have left u0 (x) blank because we must diverge for a
while in order to work it out. This is where we apply the chain
rule to make the jump between u(x) → u0 (x).
Letting w = x + 5 then,
u = w7
Now we determine what our u(x) and v(x) values are in order
to apply the quotient rule in order to check the result of the
example above.
u(x) = x2 → u0 (x) = 2x
v(x) = x → v0 (x) = 1
using the definition of the chain rule by (7),
du
du dw
=
.
dx dw dx
1.
du
= 7w6
dw
Then using the quotient rule defined in (11),
dy(x) (x)(2x) − (x2 )(1)
=
dx
x2
dy(x) 2x2 − x2
x2
=
=
=1
dx
x2
x2
as required from before. We will now look at more challenging examples.
Example
Let,
2.
dw
=1
dx
Then remembering that w = x + 5 we can finally determine
that,
u0 (x) =
du
= (1)(7w6 ) = 7w6 = 7(x + 5)6
dx
Which is required to apply the quotient rule. Now back to our
variables,
y(x) =
u(x) = (x + 5)7 → u0 (x) = 7(x + 5)6
x+4
x2
v(x) = cos(x) → v0 (x) = − sin(x)
So we define our u(x) and v(x) variables,
0
u(x) = x + 4 → u (x) = 1
v(x) = x
2
0
→ v (x) = 2x
and applying the quotient rule (11),
dy(x) (cos(x))(7(x + 5)6 ) − (x + 5)7 (− sin(x))
=
dx
cos2 (x)
Differentiation — 10/11
dy(x) 7(x + 5)6 cos(x) + (x + 5)7 sin(x)
=
dx
cos2 (x)
dy(x) (x + 5)6 [7 cos(x) + (x + 5) sin(x)]
=
dx
cos2 (x)
and we are done. As a side note, generally the quotient rule
can always be adapted to fit the definition of the product rule
by taking the denominator up to the top line and changing
power. We will reattempt the same example and solve it using
the product rule to see if we can match the two solutions.
Example
sin(x) dy(x) 7(x + 5)6
=
+ (x + 5)7
dx
cos(x)
cos2 (x)
We multiply the left hand fraction top and bottom by cos(x)
which allows us to add the two fractions together as they will
have the same, common denominator,
dy(x) 7(x + 5)6 cos(x) (x + 5)7 sin(x)
=
+
dx
cos2 (x)
cos2 (x)
dy(x) 7(x + 5)6 cos(x) + (x + 5)7 sin(x)
=
dx
cos2 (x)
y(x) = (x + 5)7 (cos(x))−1
We then define our variables again as u(x) and v(x) like we
did in the quotient rule example,
u(x) = (x + 5)7 → u0 (x) = ...
v(x) = (cos(x))−1 → v0 (x) = ...
To determine u0 (x)
We have u(x) = (x + 5)7 . Then to determine u0 (x) we must
let, w = x + 5:
u(x) = w7
Using the chain rule defined by (7),
du(x) du(x) dw
=
.
= (7w6 )(1) = 7w6 = u0 (x)
dx
dw dx
To determine v0 (x)
We have v(x) = (cos(x))−1 . Then to determine v0 (x) we must
let, z = cos(x):
v(x) = z−1
Using the chain rule defined by (7),
dv(x) dv(x) dz
=
. = (−z−2 )(− sin(x))
dx
dz dx
dv(x) sin(x)
=
= v0 (x)
dx
z2
Now we can finally write down what all of our ingredients are
to solve the problem,
u(x) = (x + 5)7 → u0 (x) = 7w6
v(x) = (cos(x))−1 → v0 (x) =
Then to simplify things we must substitute our values in for
w = x + 5 and z = cos(x),
sin(x)
z2
So using the product rule defined by (10),
dy(x)
= v(x)u0 (x) + u(x)v0 (x)
dx
sin(x) dy(x)
= (cos(x))−1 (7w6 ) + (x + 5)7
dx
z2
dy(x) (x + 5)6 [7 cos(x) + (x + 5) sin(x)]
=
dx
cos2 (x)
which is exactly the result given back in the previous example.
I think it is in agreement that the quotient rule, while it is
another rule to learn, is slightly simpler to apply in this and
most other cases.