### Video Transcript

In this video, weβll learn how to
find the derivative of exponential functions. Weβll begin by stating the formula
for the derivative of an exponential function in the form π of π₯ equals π to the
π₯ and π to the ππ₯, before applying this formula to examples of increasing
complexity. Weβll also see how we can use the
laws of exponentials and logarithms to find the derivative of general exponential
functions.

Remember, an exponential function
is one of the form π of π₯ equals ππ to the power of π₯, where π and π are
constants and π is greater than zero. Weβre actually first going to
consider the derivative of a special form of this, the exponential function π of π₯
equals π to the π₯ and π of π₯ equals π to the ππ₯. Now a common misconception is to
think that we can apply the power rule to these exponential functions. In fact, the power rule only applies
when the exponent is fixed and the bases are variable.

In the exponential functions, the
base is fixed and the exponent is the variable. So weβll instead need to use a
separate formula for the derivative of an exponential function, though the
derivation of this formula is achievable if you know how to differentiate the
inverse of π to the π₯ or ln π₯. And it can also be achieved through
first principles. But it is outside of the
constraints of this video to look at all of these. So instead, weβll state the formula
for the derivative of π to the power of π₯ and π to the power of ππ₯.

The derivative of π to the power
of π₯ with respect to π₯ is π to the power of π₯. And this formula can be
generalized. And we can say that the derivative
of π to the power of ππ₯ with respect to π₯ is ππ to the ππ₯. Now, the function π of π₯ equals
π to the π₯ is incredibly unusual. Its derivative is the same as the
original function. Geometrically, this means that, for
every value of π₯, the slope or the gradient of the tangent to the curve at that
point is equal to the π¦-value. For example, when π₯ is equal to
two, π¦ is equal to π to the power of two, which is approximately 7.39. Since the derivative of π to the
power of π₯ is π to the power of π₯, this means that the gradient of the tangent to
the curve at that point is also 7.39. Letβs have a look at an example of
the application of these formulae.

If π of π₯ is equal to negative
five π to the negative nine π₯, find π dash of π₯. Find the derivative of the
function.

Remember, the derivative of π to
the power of π₯ is π to the power of π₯. And the derivative of π to the
power of ππ₯ is ππ to the ππ₯. Now, our function is some multiple
of π to the ππ₯. Itβs negative five times π to the
ππ₯, where π is equal to negative nine. Now, we know that the constant
factor rule allows us to take constants outside a derivative and concentrate on
differentiating the function of π₯ itself. So this means we can say that the
derivative of the function of π₯ is equal to negative five times the derivative of
π to the negative nine π₯. And we know that the derivative of
π to the negative nine π₯ is negative nine times π to the negative nine π₯. And since negative five multiplied
by negative nine is 45, we can say that the derivative of our function is 45π to
the negative nine π₯.

Letβs consider another example.

Find dπ¦ by dπ₯ if five π¦π to the
two π₯ equals seven π to the five.

Now, at first glance, this does
look a little complicated. However, we can clearly see that we
can rearrange the equation to make π¦ the subject. Weβre going to divide both sides of
the equation by five π to the two π₯. On the left-hand side, that leaves
us simply with π¦. And on the right-hand side, we have
seven π to the power of five over five π to the two π₯. Now, actually, one over π to the
power of two π₯ is equal to π to the negative two π₯. So we can rewrite our equation. And we say that π¦ is equal to
seven π to the power of five over five times π to the negative two π₯.

Notice that seven π to the five
over five is just a constant. So we can differentiate this using
the general formula for the derivative of the exponential function. The derivative of π to the ππ₯
with respect to π₯ is ππ to the ππ₯. And of course, remembering that the
constant factor rule allows us to take constants outside a derivative and
concentrate on differentiating the function of π₯ itself.

In other words, dπ¦ by dπ₯ is equal
to seven π to the power of five over five times the derivative of π to the
negative two π₯ with respect to π₯. And the derivative of π to the
negative two π₯ with respect to π₯ is negative two π to the negative two π₯. This means dπ¦ by dπ₯ is negative two
times seven π to the power of five over five times π to the negative two π₯.

Notice that the derivative can
actually be expressed in terms of π¦ since we said that π¦ was equal to seven π to
the power of five over five times π to the negative two π₯. And we can therefore say that dπ¦
by dπ₯ is equal to negative two π¦.

Next, we consider a slightly more
complex example, in which weβll need to apply other rules for differentiation.

Find the derivative of the function
π of π§ equals negative three π to the four π§ over four π§ plus one.

Here we have a function of a
function or a composite function. This tells us weβre going to need
to apply the chain rule. This says that the derivative of π
of π of π₯ is equal to the derivative of π of π of π₯ multiplied by the
derivative of π of π₯. Alternatively, we can say that if π¦
is equal to this composite function, π of π of π₯, then if we let π’ be equal to
π of π₯, then π¦ is equal to π of π’. And this means we can say that the
derivative of π¦ with respect to π₯ is equal to the derivative of π¦ with respect to
π’ multiplied by the derivative of π’ with respect to π₯.

In this example, we say that π¦ is
equal to negative three times π to the power of π’, where π’ is equal to four π§
over four π§ plus one. And since weβre going to be
differentiating with respect to π§, we alter the formula slightly. And we say that dπ¦ by dπ§ is equal
to dπ¦ by dπ’ times dπ’ by dπ§. So weβre going to need to work out
dπ¦ by dπ’ and dπ’ by dπ§. dπ¦ by dπ’ is a fairly easy one to differentiate. We know that the derivative of π
to the power of π’ is π to the power of π’. So the derivative of negative three
π to the power of π’ is negative three π to the power of π’. And then, we can replace π’ with
four π§ over four π§ plus one to get negative three π to the four π§ over four π§
plus one. But what about the derivative of four
π§ over four π§ plus one?

Well, here, we need to use the
quotient rule. This says that the derivative of π
of π₯ over π of π₯ is equal to the function π of π₯ times the derivative of π of
π₯ minus the function π of π₯ times the derivative of π of π₯. And thatβs all over π of π₯
squared. We change π of π₯ to π of π§ and
π of π₯ to π of π§. Then, the derivative of the
numerator of our fraction is four. And the derivative of the
denominator of our fraction is also four. So the equivalent to π of π§ times
the derivative of π of π§ is four π§ plus one times four. And the equivalent to π of π§
times the derivative of π of π§ is four π§ times four. And thatβs all over the denominator
squared. Thatβs four π§ plus one
squared. Now, distributing the parentheses
and we simply end up with four on the numerator of this fraction. So dπ’ by dπ§ is equal to four over
four π§ plus one squared.

Letβs now substitute everything we
have into the formula for the chain rule. dπ¦ by dπ’ times dπ’ by dπ§ is negative
three π to the four π§ over four π§ plus one times four over four π§ plus one
squared. And if we simplify, we see that the
derivative of our function is negative 12π to the four π§ over four π§ plus one all
over four π§ plus one squared.

Our next example will lead us into
a definition for the derivative of an exponential equation in the form ππ to the
power of π₯.

If π¦ is equal to negative three
times two to the power of π₯, determine dπ¦ by dπ₯.

To answer this question, weβll
first recall the fact that the constant factor rule allows us to take constants
outside of a derivative and concentrate on differentiating the function of π₯
itself. We can therefore say that dπ¦ by dπ₯
is equal to negative three times the derivative of two to the power of π₯ with
respect to π₯. But how do we differentiate two to
the power of π₯? Well, weβre going to use the laws of
logarithms and exponentials.

We start by saying that two is the
same as π to the power of ln two. And remember, this is true because
π and ln are inverse functions of one another. Next, we raise both sides of this
equation to the power of π₯ and use the laws of exponents to say that two to the
power of π₯ is equal to π to the power of ln two times π₯. Well, ln two is just a
constant. So we use the fact that the
derivative of π to the ππ₯ is ππ to the ππ₯. And we can say that the derivative
of two to the power of π₯ is ln two times π to the ln two π₯. This means that dπ¦ by dπ₯ is equal
to negative three multiplied by ln two times π to the ln two π₯.

Now, remember, we actually defined
two to the power of π₯ to be π to the ln two π₯. So we replace π to the ln two π₯
with two to the power of π₯. And we see that dπ¦ by dπ₯ in this
case is negative three times ln two times two to the power of π₯.

Now, we can generalize the result
from the previous example. And we can say that the derivative
of an exponential function of the form π to the power of π₯ is ln π times π to
the power of π₯. And we can generalize that
somewhat. And we say that the derivative of
π to the power of ππ₯ is equal to π times ln of π times π to the power of
ππ₯. And whilst itβs very useful to
learn this formula by heart, itβs also important that youβre able to follow the
process that we took before for finding them. Letβs look at the application of
these formulae.

Find the first derivative of the
function π¦ equals seven to the power of negative nine π₯ minus eight all to the
power of negative two.

To find the derivative of this
function, weβll first want to see if thereβs a way in which we can simplify it
somewhat. In fact, we can use the laws of
exponents to do so. Remember, π to the power of π to
the power of π is equal to π to the power of π times π. So we can say that seven to the
power of negative nine π₯ minus eight all to the power of negative two is the same
as seven to the power of 18π₯ plus 16. We then use another law of
exponents. This time, π to the power of π
times π to the power of π is the same as π to the power of π plus π. So π¦ must be equal to seven to the
power of 18π₯ times seven to the power of 16.

Now, seven to the power of 16 is a
constant. So we can say that the first
derivative of our function dπ¦ by dπ₯ is equal to seven to the power of 16 times the
derivative of seven to the power of 18π₯ with respect to π₯. And here, we use the fact that the
derivative of π to the power of ππ₯ with respect to π₯ is equal to π times ln of
π times π to the power of ππ₯. And this means the derivative of
seven to the power of 18π₯ is 18 times ln of seven times seven to the power of
18π₯.

Now, once again, weβll use the fact
that π to the power of π times π to the power of π is the same as π to the
power of π plus π. And we can rewrite seven to the
power of 16 times seven to the power of 18π₯ as seven to the power of 18π₯ plus
16. And this means the first derivative
of our function is 18 times ln seven times seven to the power of 18π₯ plus 16. In our final example, weβll look at
a worded problem involving the derivatives of exponentials.

The radioactive decay of Radon-222
is modeled by the following formula. π of π‘ equals π nought times a
half to the power of π‘ over π‘ half, where π of π‘ is the remaining quantity, in
grams, of Radon-222 which has not decayed after π‘ days. π nought is the initial quantity
of Radon-222. And π‘ half is its half-life. A particular sample initially
contained 10 grams of Radon-222. Given that the half-life of
Radon-222 is 3.8215 days, find the rate of decay of the sample 10 days later. Give your answer to three
significant figures.

Remember, rate of change or, here
rate of decay, will always correspond to the gradient function or the
derivative. To answer this question then, weβre
going to need to differentiate the function π nought times a half to the power of
π‘ over π‘ half with respect to π‘. Now, this may look a little
tricky. However, π nought is a constant as
is π‘ half. And we use the fact that the
derivative of π to the ππ₯ is π times ln π times π to the ππ₯. And we can say that the derivative
of π with respect to π‘ is equal to π nought times one over π‘ half, since in our
equation thatβs the equivalent of π, times ln of a half, since in our equation π
is a half, times a half to the power of π‘ over π‘ half.

Now, we rewrite ln of a half and say
that thatβs the same as ln of two to the power of negative one. And we use the laws of
logarithms. And we see that ln of a half is the
same as negative one times ln two or just negative ln two. And so we can rewrite our
expression a little. Now, since a particular sample
initially contained 10 grams of Radon-222, we can say that π nought must be equal
to 10. We know that π‘ half is equal to
3.8215. And weβre looking to find the rate
of decay when π‘ is equal to 10. So weβll substitute all of these
into our equation for the derivative. And we get 10 over 3.8215 times
negative ln two times a half to the power of 10 over 3.8215. And that gives us a value of
negative 0.2957.

But what does this actually mean
with respect to our question? Well, correct to three significant
figures, the rate of change is equal to negative 0.296. Now, since this is negative, it
means the sample is decaying at a rate of 0.296 grams per day.

In this video, weβve seen that the
derivative of π to the power of π₯ is π to the power of π₯. And weβve seen that this can be
generalized. And we can say that the derivative
of π to the power of ππ₯ is equal to π times π to the power of ππ₯. We also saw that we can
differentiate π to the power of π₯ with respect to π₯ to get ln π times π to the
power of π₯. And we generalize that to say that
the derivative of π to the power of ππ₯ is equal to π times ln π times π to the
power of ππ₯. We also saw that it can be useful
to simplify or manipulate expressions using the rules of exponents before
differentiating. And we said that itβs important not
to confuse the power rule and the rule for differentiating exponential
functions.