Welcome to a video on the derivative of an exponential function. The goal of this video
is to differentiate exponential functions. Let's go and take a look at the rules. If our base is e,
the derivative of e to the x is equal to e to the x. This is actually quite amazing!
This function is its own derivative, meaning if we evaluate the function at a given value of x,
the function value also tells us the slope of the tangent line at that x value.
The second formula here just applies the chain rule. If the exponent is
some function in terms of x, we would apply the chain rule, and the derivative
would be e to the u times u prime. Now, if the base of the exponential is not e,
then notice the derivative would have an extra factor of natural log a times a to the x. And
then again, this last rule is if we have a raised to some function in terms of x,
we would have to apply the chain rule, so we'd have natural log a times a to the u times u prime.
Let's take a quick look at a rough proof of the derivative of the function f of
x is equal to e to the x. We start with the basic exponential function,
apply the limit definition of derivative, do our substitution, do some fancy algebra here,
factor out e to the x, and what we have is e to the x times this special limit. What we're
going to argue now is if we knew the value of a that would make this limit equal to one,
then we would have the case where the exponential function is its own derivative.
So what we do is we take that special case over here: what would a have to be in order
for this limit to equal one? Well, if we assume it's equal to one, solve it for a to the h,
and raise both sides of the equation to the power of one over h, we'd have a equal
to this expression here, which we could then conclude that if this limit is equal to one,
then a would be equal to this limit here, and by definition, this limit is equal to e. Therefore,
we have a equal to e. So when a is equal to e, going back over to the left side,
we have the derivative of e to the x is equal to e to the x.
Now, you may want to pause and take a closer look at this proof, but that's the general
idea. Now the next cases: what do we do if the base isn't e? Well, from some of our properties,
we can rewrite a to the x as e to the power of x times natural log a, and in this form,
we can just apply the chain rule to the derivative of e to the u. So what we're saying here is that
the derivative of a to the x is equal to this derivative. If we let this equal e to the u,
the derivative of this function would be e to the u times u prime. Now, natural log a
is just a constant, so the derivative of u is just one times natural log a,
which gives us natural log a times a to the x when a does not equal e.
Again, you may want to pause the video to take a closer look at this, but that's the general idea.
Let's go ahead and put it into practice. Let's say we want to find the derivative of each of these
functions. So the derivative of this function, to find the derivative of this function, we are going
to need this derivative formula: e to the u times u prime, where our value of u is equal to 6x.
So first we have the derivative of this term, which would be 10x
to the 4 minus 3. Here’s our formula: e to the 6x times u prime. The derivative
of 6x would be 6. We can clean this up to 10x to the 4 minus 18e to the 6x.
Now number two, we have to realize that this is
a product of two functions. We'll call the first function f and the second function g,
so we're going to have to apply the product rule. And so what I've done already is set
the derivative of the first function times the derivative of the second plus the second times
the derivative of the first. Let's go ahead and find these derivatives and then simplify.
Now the derivative of e to the -2x, we are going to have to apply the chain rule again,
so our derivative is e to the u times u prime. So we'd have e
to the -2x times -2 plus e to the -2x times the derivative of x cubed. That
would be 3x squared. We have -2x cubed e to the -2x plus 3x squared e to the -2x.
Now I will factor this one. We can see these two terms have a common factor of
x squared e to the -2x. Now, if we factor this out of the first term, we'd be left with -2x.
If we factored x squared e to the -2x out of this term, we'd have a positive 3. So I'll
put the positive 3 in and the -2x squared. And now there's our derivative in factored form.
Let's go ahead and try another one. Here we have a quotient. So on the right side here,
I wrote out the quotient rule. Remember it's low d high minus high d low, low low. And over here
on the left, I've already set it up. We have the denominator times the derivative of the numerator
minus the numerator times the derivative of the denominator all over the denominator squared.
The derivative of e to the 3x would be e to the 3x times u prime,
so there’s 3 minus e to the 3x times 6x to the 5 all over x to the 12th. Let’s clean this up and
then see if we can factor it. Now I notice the numerator has a common factor of 3x to the 5th
e to the 3x. Let’s factor that out, so we'd be left with x minus 2 over x to the 12th.
We can see we have an x to the 5th here and an x to the 12th here. This would simplify,
and this would become x to the 7th, so our final derivative would look like this.
Okay, let’s try another question. Determine the slope of a tangent line to the function
at the given point. What we know in order to find the slope of the tangent line,
we have to find the derivative of this function. Let’s go ahead and
do that. It would be 2 times the derivative of e to the 3x.
Again, hopefully we’re pretty good now recognizing that our u would equal -3x, so we’re going to have
e to the u times e to the -3x times -3. So our derivative function is -6 e to the 3x. If we
want to find the slope of the tangent line at this point, we need to evaluate this function
at x equal to 0. Of course, e to the 0 would be 1, and -6 times 1 gives us a slope of -6.
Let’s go ahead and graph this function and check our work. So here’s our function;
here’s our point of tangency. The question is: does this line have a slope of -6? If we start
up here and go down six units and right one, we have the correct line, which verifies our work.
Let’s go ahead and take a look at one more application question. More Americans are buying
organic food and vegetables made with organic ingredients. The amount A of in billions of
dollars spent on organic food T years after 1995 can be approximated by this function.
We have two questions to consider: estimate the amounts that Americans will spend on organic food
in the year 2009. So the question is, what value of T are we going to use? Well, based upon the
question, 1995 was our base year, so we have to take our desired year and subtract 1995. This will
give us 14, so this is not a calculus question; we just need to evaluate A at a value of 14 for T.
Second, natural log brings up the base e. We have approximately 3.2. Now if
you recall from the question, it was 30.2 billion dollars. A of F was in billions.
And the last question: estimate the rate at which the spending
on organic food was growing in the year 2006. Whenever we hear that word "rate" or "rate of
change," we should be thinking derivative, so we need to evaluate the derivative. Let’s
start by finding our value of T that’s going to give us a T value of 11 for this question.
But on this one, we want to find the rate of change; we need to find the derivative.
Remember the derivative of this would be e to the u times u prime. Of course,
u prime would be 0.18. So simplifying this, we would obtain the derivative
function here. So if we simplify this, we’d obtain this derivative function.
And now we need to evaluate this at T equals 11, which gives us approximately 3.17. Again,
this is in billions of dollars, but it's a rate,
so billions of dollars and time is in years, so per year at that instant where T is equal to 11.
I'm going to go back to the graphing calculator just for a moment and show you another way you
can check your work. This graphing calculator will evaluate derivatives at a specific value of x or,
in this case, T. So if we hit our math key and scroll down to option 8, If we type in the
original function, now I’m going to use x instead of T, then type in comma x, comma the value of x
we want to determine the derivative at, which is 11. It will give us the same answer. It's a
nice way to check, but I'm sure your instructor will want you to show the derivative function.
Okay, I hope that helps explain how to find the derivative of
exponential functions. Thank you for watching.