Professor Dave here, let’s take some derivatives.
In the previous tutorial, we made the connection
between the concept of differentiation, and
the act of taking the derivative of a function,
and we even derived the power rule. Being
able to connect all of these concepts will
give us a dramatically enhanced understanding
of what it is to do calculus, and why we need
to do it. But the fact of the matter is, there
are times where we simply have to know an
algorithm, and be able to use it to get correct
answers. So whether or not the previous tutorials
were of help to you, it’s now time to practice
using the power rule to take the derivative
of various polynomial functions, something
that every calculus student will absolutely
have to be able to do.
As we said, the power rule tells us that in
order to take the derivative of some function
involving a positive integer exponent, we
just pull the exponent down to become a coefficient,
and then reduce the exponent by one. So the
derivative of x squared is two times x to
the one, or simply two x. At this point, let’s
introduce another type of notation. When we
write d over dx, this notation implies differentiation.
It means that we are differentiating something
with respect to x, which is the same thing
as taking the derivative of something, so
just as f prime of x means the derivative
of f of x, d over dx followed by some function
means to take the derivative of that function
with respect to x. So in short, f prime of
x is completely identical to d over dx, f
of x. Sometimes we will use one notation,
sometimes we will use the other. With that
understood, let’s take the derivative of
some more polynomials. Again, using the power
rule, the derivative of x cubed is three x
squared. The derivative of x to the fourth
is four x cubed. This pattern will continue
for any exponent. Now let’s take the derivative
of three x to the fourth. As we are bringing
this four down here, we must multiply by the
three that is already there, so we get twelve
x cubed. Another way of explaining this is
by mentioning the constant multiple rule,
which says that the quantity (CF) prime equals
c times f prime. If we are taking the derivative
of some function that is being multiplied
by some constant, this will be the same as
multiplying the constant times the derivative
of the function, so we can just pull the constant
out here. Using this to look at the previous
example, the derivative of three x to the
fourth, we can pull the three out, take the
derivative of x to the fourth to get four
x cubed, and then we simply multiply by the
three that we pulled out before, and we get
the same answer that we already had. So however
we think about it, when taking a derivative,
the exponent that we pull down front must
multiply any existing coefficient.
Now what about polynomials with multiple terms?
Actually, this will be no problem, due to
the sum rule. This says that the derivative
of the quantity f of x plus g of x, will be
equal to the derivative of f of x plus the
derivative of g of x, or more simply, (f plus
g) quantity prime equals f prime plus g prime.
Similarly, there is the difference rule, which
works just like the sum rule, stating that
(f minus g) quantity prime equals f prime
minus g prime. That means that no matter how
many terms are in a polynomial, we will just
take the derivative of one term at a time.
So for (x to the sixth) plus (two x to the
fourth) minus (three x squared) plus two,
we can just go from left to right to get (six
x to the fifth) plus (eight x cubed) minus
(six x). Don’t forget that the two will
go away as it’s a constant. So as long as
we know the power rule, taking the derivative
of any polynomial in standard form is actually
quite trivial. Let’s try just one more for
practice. Try (x to the eighth) minus (five
x to the fifth) plus (eleven x cubed) minus
(two x). Again, moving left to right, we get
(eight x to the seventh) minus (twenty five
x to the fourth) plus (thirty three x squared)
minus two.
Now with the power rule under our belts, let’s
try something a little trickier. What about
the derivative of a product of functions?
Unfortunately, this is not as simple as just
the product of the two derivatives. Inotherwords,
fg quantity prime is not equal to f prime
times g prime. Instead, we have to use the
product rule. This says that the derivative
of f of x times g of x is equal to (f of x)
times (g prime of x) plus (g of x) times (f
prime of x), or more simply, fg quantity prime
equals f, g prime, plus g, f prime. We can
even verify that this works by looking at
x squared. We know the derivative is two x,
but say we turn this into x times x, and apply
the product rule. X times the derivative of
x is simply x, and the derivative of x, times
x, is also x, so we add them together to get
two x, and we can see that this works just
as expected. Let’s try this with some other
functions. What is the derivative of (x cubed)
times (two x)? Well first, let’s leave the
first function alone and take the derivative
of the second. The derivative of two x is
two, so we get two x cubed. Now, let’s take
the derivative of the first function and leave
the second one alone. That will be three x
squared, times the two x that we didn’t
touch, and that will be six x cubed. Adding
them up, we get eight x cubed. As we can see,
this is not simply the product of the derivatives
of each function, we had to use the product
rule, and if we had simply multiplied these
two original terms together to get two x to
the fourth, the power rule would have given
us the same result, eight x cubed. Let’s
try one more. Find the derivative of three
x to the fourth times five x squared. We can
use precisely the same algorithm every time,
so let’s leave the first function alone,
and take the derivative of the second to get
ten x, which gives a product of thirty x to
the fifth. Then we take the derivative of
the first to get twelve x cubed, and leave
the second function alone, which then multiply
to give sixty x to the fifth. Adding them
together, we get ninety x to the fifth. Again,
this agrees with the answer we would get if
we were to multiply the original terms together
and use the power rule. Later, we will see
plenty of functions where we can’t just
multiply these terms together, making the
product rule absolutely necessary. So as long
as we can memorize the product rule, which
is fairly simple, it’s just a matter of
taking a couple of simple derivatives, followed
by some multiplication and addition.
Lastly, what about the derivative of a quotient
of functions? Say we want to find the derivative
of f of x over g of x. To do this, we need
the quotient rule, and unfortunately, this
is not an exact parallel of the product rule.
This is a different rule that should also
be memorized. To find the derivative of f
of x over g of x, we find (g of x) times (f
prime of x), minus (f of x) times (g prime
of x), and all of that will be over g of x
quantity squared. Again, more simply, f over
g quantity prime equals g f prime minus f
g prime all over g squared. Make note that
for the product rule, the order of the terms
didn’t matter, since it involved a sum,
but for the quotient rule it will matter.
It must be g times f prime minus f times g
prime. If you switch this, you will get incorrect
answers, and we also can’t forget the g
of x quantity squared on the bottom. Don’t
worry, you will memorize this quite easily.
Let’s try an example. What about the derivative
of (x squared plus 1) over (x plus two)? First
let’s set up the expression, making sure
to put everything in the correct place. Then
let’s evaluate these two derivatives. For
the first, we get two x, and for the other
we get one. Now we can multiply these terms,
distribute this negative sign and simplify,
and expand the denominator if we wish, to
get (x squared plus four x minus one) over
(x squared plus four x plus four), which can
be left as is.
So these are the basic algorithms we will
use in order to take the derivatives of functions.
We will expand on them as we go, finding that
when using the power rule, this exponent can
be a negative integer, or even a non-integer,
but the rules don’t really change, and even
when products or quotients get more complicated,
it’s not really any harder, just more steps,
and therefore more opportunities to make a
mistake. But if we practice and become diligent,
we will find that it’s not too difficult
to take the derivative of any polynomial.
Let’s check comprehension.
