 Hi everyone, and welcome to this lesson on derivatives of exponential functions. We'll be walking through a couple examples as well as an application problem at the end, but we're going to begin with a calculator exploration. So if you grab your calculators, we're going to start out by really discovering what the derivative of the function y equals e to the x equals. So under y equals, if you just set that up to graph e to the x, and we'll just do a standard graph, zoom six, and of course we all know that's what it looks like. So in order to discover what the derivative of e to the x is, we're going to take note of a couple different values, function values at various x values, and then we're going to use the calculators built in functions to calculate the derivative at those same x values. So let's start with determining the function value when x equals two. So when x equals two, the function value is approximately 7.389. So let's determine next the derivative, the slope of the tangent line, at that same x value. Remember you can do that graphically with the built-in features down at number six, dy dx, and simply type in two and enter, and notice how the derivative at that point is that same 7.389. Interesting. So let's try another one. Let's try negative one. So let's begin with the function value at negative one, and we get approximately 0.368, and now we're going to figure out the slope of the curve, the derivative at that same point. Down to number six again, and simply type in negative one and enter, and there is that same 0.368. So we're two for two. Let's try another one. Let's try when x equals a half, and we'll begin again with the function value. The function value when x equals a half is 1.649. So let's now do the derivative at that same x value, and once again we get 1.649. We could sit here all day and do this with different x values, and you will notice that the derivative value of e to the x is always equal to the function value at e to the x for that same x value, which really leads to the rule. The derivative of e to the x is simply going to be e to the x. It's the only function which is its own derivative, kind of interesting. Now of course you are going to run into situations with the chain rule. With the chain rule, as we learned with other functions, it's going to be the same base function, so you'd have e raised to the exponent, but then you simply need to multiply by the derivative of that exponent. So we'll be walking through some of those as we do our examples today. So next let's address the derivative of a to the x. So we're going to assume that a is a positive real number, not equal to 1, obviously because if we had 1 to the x, that's simply going to be 1. The derivative rule is that it's a to the x multiplied by the natural log of a. Now if you think about this rule and then apply it to the one we just looked at with e to the x because e is a positive real number, if you go back to this rule and apply that a to the x rule to here, remember what we would have is e to the x, but then you're multiplying by natural log of e, which we just know to equal 1 anyway. So the rule does actually bear out. The proof for this is generally straightforward. It involves taking the derivative of both sides of the equal sign after you rewrite once you have it as y equals a to the x. It actually does work out. Let me see if I can give you a mini proof over here. So if we have y equals a to the x, if we just take the natural log of both sides, something you learned to do when you solved exponential equations, of course we can pop the x out in front, so it looks like that. By doing the derivative of both sides, derivative of natural log of y is going to be 1 over y. You might not have learned that yet. Multiply by y prime. You have to remember this natural log of a is a number, which is really the coefficient of the x. So the derivative over here is simply natural log of a. When you go to solve for y prime and multiply both sides by y, well y is equal to a to the x. So it's a relatively straightforward proof. Once again, though, you'll run into occasions with the chain rule. You'll have the same basic rule, so you'll have a to the u times natural log of a, but then you just need to multiply by the derivative of the x prime. So let's take a look at a few examples. And all we're doing is doing straightforward derivatives. So with the first one, of course, we're going to rewrite that square root function as x to the 1 half and think of it that way. So our derivative of the square root function, you've already learned to do that with the power rule. Now with the other one, remember the three stays put for now. Now we have to do the derivative of e to the 4x. So that's going to be e to the 4x. But then we have to multiply by the derivative of the exponent, which is just 4. So in the end, our answer is 1 half x to the negative 1 half minus 12 e to the 4x. So the next one involves the other rule. So to start, the 4 will stay put. When we go to do the derivative of this, it's going to be 10 to the x. Derivative of x is, of course, just 1, so no need to write that down, times natural log of 10 minus, and of course this is just a power rule situation. So you could leave it like that. Sometimes you'll see the 4 and the natural log of 10 kind of brought together as a coefficient of that 10 to the x. Otherwise that would really be all you could do. Now in this one, remember that e and pi are both numbers. So when we go to take the derivative of e to the pi, that really is just a number. That derivative is going to be 0. The derivative of pi raised to the 2x, that's really the rule for a to the x, or a to the u, because it would be a chain rule problem. Pi, of course, is a positive real number. So we're going to have pi raised to the 2x times the natural log of pi, but then we need to multiply by the derivative of the exponent, which is just 2. So you could maybe see it rewritten as 2 natural log of pi. That's sort of like a coefficient of sorts times pi to the 2x. Now in the next one, you do need to remember, again, natural log of 3 and natural log of 4 are both numerical values. So when we go to do the derivative of natural log of 3 times x squared, that's really just a power rule problem. So we'll have 2 natural log of 3, and that's just times x, plus we'll keep the natural log of 4 there for now, and now we're doing the derivative of that. So that's that e to the u rule. So we'll have e to the negative 9x times negative 9. So we can rewrite it a little bit. Add 2 natural log of 3 times x. We can make this minus 9 natural log of 4, perhaps, e to the negative 9x. A couple more before we get to our application problem. So for this one, let's think of it as e raised to the x to the 1 half, kind of a super chain rule problem. So think of it as the rule for e to the u. It is a chain rule. So we'll have to keep the function as it is. So e, if you prefer to write it at this point, is just square root of x. That's fine, because we're essentially keeping the function. But now remember we need to multiply by the derivative of that exponent. So that's where we're taking the derivative now of the x to the 1 half. So it's perfectly fine to leave it like that. If you want to rewrite it, we could have e to the square root of x over 2 square root of x like that. Otherwise it's okay to leave it like it was. And number 6 will be another souped up chain rule problem. So think of it as just doing the derivative of 3 to the u. So we'll keep the function as is. Multiply by natural log of the base. Now we need to multiply by the derivative of the exponent. So that'll be 2x minus x. Not a whole lot you can do to rewrite... Oh, minus 1. So let's take a look at an application problem. And I'm sure you've studied exponential functions and application problems. So this is one of those. So the value of a car purchased back in 1997 can be approximated by the function v of t equals 25 times 0.85 raised to the t. Where t is the time in years from the date of purchase and v is the value of the car in thousands of dollars. So the first thing we're asked to do is simply to evaluate and interpret v of 4. So if we want v of 4, we can simply do that on our calculators. I'll give you a second to do that. So when you go ahead and do that, you should come out with about 13.050. So that's what v of 4 is equal to. In terms of the interpretation, remember this is going to represent the value, but it's in thousands of dollars. So if you multiply this point 13.050 times 1,000 and maybe even preserve more decimals that you might have had there on your calculator. What we find is that the car's value four years after purchase, approximately 13,050. And check the decimals on your calculator. It should be about 16 cents. So that would be the interpretation. So I'm sure you've done that kind of problem before. So let's get more into the calculus then. The next thing we're going to do is find an expression for v prime of t and include what the units of measure would be. So if you go back to the function, what it was, I can rewrite it here for us. So remember v of t was given to be 25 times 0.85 to the t. So when we find v prime of t, the 25 is the coefficient simply remains. Then the 0.85 raised to the t, that's like doing the derivative of a to the t, a to the x. So we're going to keep the 0.85 to the t. And we need to multiply by the natural log of 0.85. And of course the derivative of t is simply 1, so we don't really need that. Now in terms of the units of measure, remember that a derivative is a rate of change. And we know a rate of change is going to be change in y over change in x, like any slope would be. And if you think about what the y's, the v's stand for in this case, it's the cost of the car, which of course we're going to measure in dollars. And it's over the number of years from the date of purchase. So that's going to be years in the denominator. So the units for this would be dollars per year. So simply think of it as a rate of change, a slope, in order to determine what the units of measure would be. So finally let's end by interpreting and finding v prime of 4. So we have 25, you can do this of course on your calculator. So if you go ahead and do that, you should get that it's a negative approximately 2.121. So first of all, the negative, the fact that this is negative, and remember what the derivative represents in this case, it's the amount per year that the car's value is changing. So the fact that it's negative tells us the car's value is decreasing, which makes sense. The older the car gets, it's worth less money. And this is really a measure of how much it's decreasing each year. So at 4 years from the date of purchase, the car's value is decreasing. That's because it's negative, approximately. And remember this is in thousands. So again if you multiply this by a thousand and maybe even go back to your calculator value with the extra decimals, perhaps take a peek at that and it should be about $2,120.90 per year. Alright, notice how the negative is not included there. We have the negative accounted for in the fact that it's decreasing. So hopefully this has provided a good base for you in how to do derivatives of exponential functions, including an example of how you might see them with an application.