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Class Notes Calculus I

Derivatives of Exponential Functions; Product and Quotient Rules

On today's quiz we were to determine whether we could take a derivative of each of several expressions with our present knowledge, consisting of our ability to take derivatives of power functions, exponential functions, sums and differences of functions, and constant multiples of functions.

The functions and our results are summarized below.

The first function, (4 t^2 - 7 / t^5) / t^4, is easily enough reduced to a sum of power functions, so we can certainly take the derivative.

The second function has 2 ^ (t^2) as a term. We can certainly take the derivative of a power function or an exponential function, but we recognize this function as a composite of a power function and an exponential function and we have not yet learned how to take such a derivative.

The third function is clearly a difference of power functions, so we can take its derivative. The same is true of the fourth function.

The fifth function, (4 t^2 - 5) / (2t^2 - 4), is a rational function with the polynomial has numerator and another polynomial as denominator. Both the these functions have real zeros but they do not coincide so we cannot reduce the rational function to power functions. We therefore cannot take the derivative (at least not until later today when we learn the rules for product functions in quotient functions).

The same is true of the sixth function.

Applying the rules of derivatives.

The last two functions are sums of power functions and exponential functions and we can easily take their derivatives.

Derivatives of sums of power functions and exponential functions easily found using the sum and constant-multiple rules.

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In response to questions from the class, we note that 7 ^ (`sqrt(t)) is a composite of a square root function and an exponential function. We will soon be able to take the derivative of such a function, but we will require the chain rule to do so.

The derivative of x^3 + 3^x is the sum of the derivatives 3x^2 of the power function x^3 and the derivative ln(3) * 3^x of the exponential function 3^x.

We quickly review the process of deriving the formula for the derivative of the e^x function. The essential step requires that we know that for small values of h, e^h = 1 + h is a good approximation. During the last class we stop how this approximation arises from the definition of the number e and the binomial formula. Armed with this approximation we easily write down the definition of the derivative of e^x and, aided by the laws of exponents, refunded to derivative is e^x. The sure you can reproduce this derivation when asked.

7 raised to the power square root of t is the composite of a power function and an exponential function. Its derivative will be found using the chain rule. x^3 + 3^x is the sum of a power function and an exponential function; apply the sum rule.

A text problem asks that we find the equation of the line tangent to y = 1 - e^x at the point for the graph of the function crosses the x axis. We construct a graph of this function and quickly realize that it crosses the x axis at the origin (0,0).

To find the equation of the tangent line at this point, we need only find the slope of the graph at this point. The slope is the derivative. The derivative of y = 1 - e^x is easily found to be y' = - e^x; and x = 0 we therefore have slope y' = -e^0 = -1.

1 - e^x crosses the x axis where 1 - e^x = 0, so e^x = 1 and x = 0. Its derivative is -e^x; at x = 0 the tangent line has slope -e^0 = -1.

We can now construct the tangent line. The sketch below shows the graph of the function and the point (0, 0). We sketch the tangent line in red and note that its slope is -1. We then place an arbitrary point (x, y) on this tangent line and note that the slope from (0, 0) to (x, y) is rise / run = (y-0) / (x-0) = y / x. Since the slope is -1, we therefore have y / x = -1 so that y = -x.

The tangent line to 1 - e^x where the graph crosses the x axis is a line with slope -1 through the point (0 , 0), so the tangent line is y = -x.

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The Product and Quotient Rules

Suppose we wish to take the derivative of a function like y = x^2 * e^x. We can certainly take the derivative of x^2, and we can also easily take the derivative of e^x. However, as we have seen repeatedly, we as yet have no means of taking a derivative of a product of two functions. We might be tempted to simply assert that the derivative is the product of the derivatives of the two functions, but we have no means of justifying this step, and we would in fact be seriously in error if we did so.

We see that it would be a good idea to have a means of taking the derivative of product of two functions, so we pose the problem as follows: If y(x) = f(x) * g(x), then if f(x) and g(x) have derivatives, what is the derivative y'(x) = ( f(x) * g(x) )' ?

We can easily write down the definition of this derivative as the limit of its difference quotient:

y'(x) = lim [`dx -> 0] { (f(x + `dx) * g(x + `dx) - f(x) g(x) ) / `dx }.

That looks fine, but how the world do we simplify that numerator?

We might first note that we can represent the product f(x) * g(x) as the area of a rectangle with sides of length f(x) and g(x). This rectangle is pictured at upper right in the figure below, and it is noted that the area of this 'red' rectangle represents one of the terms in the numerator of the difference quotient.

How do we find the derivative of y = x^2 e^x? We develop the rule for taking the derivative of the product of two functions.

Now if f and g are continuous functions, as they must be if they have derivatives, then f(x + `dx) and g(x + `dx) are not much different from f(x) and g(x). The differences might be positive or negative. For the purposes of the figure below we will assume that the differences are positive, though we could easily enough adapt the picture to a situation for one or both differences are negative.

We therefore expand the 'red' rectangle to form a 'purple' rectangle whose dimensions are f(x + `dx) by g(x + `dx) and whose area is therefore f(x + `dx) * g(x + `dx). This rectangle contains the 'red' rectangle plus the area of shaded region. The difference between the areas is the area of the shaded region and this area therefore represents the numerator of the difference quotient.

The product f(x) * g(x) can be represented by a rectangle with dimensions f(x) and g(x). f(x + delta x) * g(x + delta x) is represented by an analogous rectangle. The difference in the areas, divided by delta x, is the approximate derivative.

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We will call the difference between f(x) and f(x + `dx), represented by the increase in width between the red and purple rectangles, `df. We will similarly call the difference between g(x) and g(x + `dx), represented by the increase in height between the red and purple rectangles, `dg. The difference between the area of the red and purple rectangles is therefore represented by three rectangular regions, one whose dimensions are `dg by f(x), another whose dimensions are `df by g(x), and a third small area whose dimensions are `df by `dg. The total area of these regions is f(x) `dg + g(x) `df + `df * `dg.

The difference between g(x) * f(x) and g(x + delta x) * f(x + delta x) is equal to f(x) * delta g + g(x) * delta f, plug delta f * delta g. For small delta x the last is tiny and becomes negligible.

The area f(x) `dg + g(x) `df + `df * `dg represents the numerator of the difference quotient. Noting that `df and `dg approach 0 as `dx approaches 0, we see that the first two terms in this numerator also approach 0 and the third term approaches 0 much faster. As we will see this last term will therefore become insignificant when we take the limit of the difference quotient.

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We now proceed to simplify the difference quotient. The first step in the simplification is to replace the numerator with the expression we have obtained from the rectangles. The second step consists merely of algebraic simplification.

To justify the third step, we note that in the limit of f(x) `dg / `dx, f(x) does not involve `dx and can therefore be considered constant for the limiting process. It follows that lim [`dx -> 0] ( f(x) `dg / `dx ) = f(x) lim [`dx -> 0] ( `dg / `dx ). Recalling that `dg = g(x + `dx) - g(x), we see that the limit must just be g'(x), so that the first term gives us f(x) g'(x).

Similar observations show that the limit of the second term must be g(x) f'(x).

The third term `df `dg / `dx could be regarded as approaching `df g'(x) as `dx -> 0; since g'(x) is bounded and `df -> 0, it should be clear that this term approaches 0.

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We can write our result as (d / dx) (fg) = f g' + g f', as above, or simply as (f*g) ' = f' g + g' f = f' g + f g', as below.

We can then use the product rule and a little ingenuity to obtain the quotient rule (f/g)' = (f' g - f g') / g^2, indicated below. Note that the numerator of the expression for the quotient rule is much like the expression we get for the product rule, except for the - sign.

If we now wish to determine the derivative of the function y = x ^ 2 * 2 ^ x, we first note that y is the product of the functions f(x) = x^2, with derivative f' = 2x, and g(x) = 2^x with its derivative ln(2) * 2^x. The derivative y' will therefore be f' g + f g' = (x^2)' * 2^x + x^2 (2^x)' = 2x * 2^x + x^2 * ln(2) * 2^x, as shown and simplified below.

The product rule and the quotient rule. Product rule applied to y = x^2 * 2^x.

We can apply the quotient rule to the function y = x ^ 2 / 2 ^ x. The functions f and g will be the same as above, but we will be applying the quotient rule. We obtain y' = (f' g - f g') / g^2, as shown below for the specific functions.

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