Professor Dave again, let’s take a few more
derivatives.
Now we know how to take the derivative of
polynomials, trig functions, products, quotients,
and even composite functions.
So let’s extend our skills of differentiation
to just one more class of functions, those
being logarithmic and exponential functions.
If you’ve forgotten what logarithms are,
go back and check out my three-part series
on this topic now, otherwise let’s do some
calculus with these functions.
Rather than going through a lot of derivations,
let’s just highlight a handful of rules
we can count on.
First let’s recall that the natural log
is the logarithm with base e.
As it happens, the derivative of the natural
log of x is equal to one over x.
From this, we can also know the derivative
of the log of x with some base a.
As we know from the change of base formula,
this can be expressed as the natural log of
x over the natural log of a, and this will
be easy to take the derivative of, because
one over the natural log of a is a constant,
so let’s pull that out, and as we said,
the derivative of the natural log of x is
equal to one over x, so putting our constant
back in, we get one over (the natural log
of a times x).
So let’s memorize these two rules for finding
the derivative of the natural log of x, and
the derivative of any logarithm of x with
some other base.
Now let’s see how we can combine these definitions
with the other rules we have learned.
Say we want to differentiate the natural log
of sine of x.
This is a composite function, so we need the
chain rule.
First we take the derivative of the natural
log, and that gives us the reciprocal of whatever
the natural log was operating on, so we get
one over sine x.
Then we multiply by the derivative of what
was inside, and the derivative of sine x is
cosine x, so we end up with cosine x over
sine x, which is equal to cotangent x.
This means that when we take the derivative
of the natural log of any function, like the
natural log of g of x, the answer will be
the derivative of what’s inside the log,
over what was originally inside, or g prime
of x over g of x.
Now what about finding the derivative of log
base ten of x squared.
Again, we need the chain rule.
We know that the derivative of the log of
base ten of something will be one over (the
natural log of ten times whatever was in the
parentheses), so we get one over (the natural
log of ten times x squared).
Now we have to multiply this by the derivative
of what was inside, and that will give us
two x, so we get two x over (the natural log
of ten times x squared), which simplifies
to two over (the natural log of ten times
x).
This can certainly get more complicated, but
our approach will always be the same, applying
the chain rule in whatever manner necessary.
Now let’s look at exponential functions.
Remember that an exponential function, rather
than something like x squared, is one where
the exponent is the variable, so swapping
these, we have two to the x.
How can we get the derivative of this?
Well the rule is that the derivative of some
exponential function, A to the x, with base
A, will be equal to the natural log of A times
A to the x.
So in the case of two to the x, we get natural
log of two times (two to the x).
The only tricky thing here is that if the
exponent is not simply x, but is rather an
expression containing x, we will then have
to multiply our answer by the derivative of
that expression.
So let’s say we have two raised to the quantity
(three x plus one).
If we take the derivative of this, just like
before we have the natural log of two times
the original function, two to the quantity
(three x plus one), but now we multiply by
the derivative of this exponent, which is
three, and we can then put the three up front.
Let’s point out one more thing now.
We mentioned that the derivative of A to the
x is the natural log of A times (A to the x).
So what if we take the derivative of e to
the x?
That would give us the natural log of e times
(e to the x).
But the natural log of e represents the power
that e must be raised to in order to get e,
and that power is one, so the natural log
of e is equal to one, and we are left with
simply e to the x.
That’s right, the derivative of (e to the
x) is (e to the x).
Rather than simply marvel at this fact, let’s
look at a fun way to prove that this is true.
When we learned about series, we found an
interesting way to represent e, by using factorials.
If we want to represent the function (e to
the x) in the same way, we can just tweak
this series a little bit, putting in the necessary
x terms, so that we get one, plus x, plus
x squared over two factorial, plus x cubed
over three factorial, all the way down the line.
So we can say that this series is equal to
e to the x.
Now let’s see what happens when we take
the derivative of this entire series, just
using the simple power rule that we know.
The derivative of one is zero, so that goes
away.
The derivative of x is one.
The derivative of x squared is two x, and
two factorial is two times one, so the two’s
cancel, and we get x.
The derivative of x cubed is three x squared,
and three factorial is three times two times
one, so the three’s cancel, and we get x
squared over two times one, which is two factorial.
For x to the fourth, again we get four x cubed,
the fours cancel, and we are left with x cubed
over three factorial.
Amazingly, this will continue all the way
to infinity, and as we will notice, the series
ends up completely unchanged.
So this is one way of proving that the derivative
of (e to the x) is simply (e to the x).
There are many other elegant ways to rationalize
this fact, but all we need to know for now
is that this function is equal to its derivative,
which means that they will be exactly the
same on a graph.
If f and f prime are the same graphically,
this means that every point on the function
is equal to its own slope at that point.
This fascinating truth will come back when we learn about integration, but for now,
let's check comprehension.
