section 5.2: derivative of the general exponential function, y = b x

Section 5.2: Derivative of the General Exponential Function, y = bx

Dana, Dina, Isabella

Consider the function f(x) = 2x.x f(x) f’(x) f’(x)/f(x)

-2 ¼ 0.1732868 0.69

-1 ½ 0.3465736 0.69

0 1 0.6931472 0.69

1 2 1.3862944 0.69

2 4 2.7725889 0.69

3 8 5.5451779 0.69

As we can see on the table, f’(x)/f(x) is constant, and equal to approx. 0.69 (less than one)… but

what does this mean?

Graphing f(x) and f’(x):

• The derivative of f(x) = 2x is equal to 0.69 x 2x.

f(x) = 2x

f’(x) = 0.69 • 2x

As you can see on the graph, f’(x) is a vertical compression of f(x) by a factor of 0.69.

Consider the function f(x) = 3x.x f(x) f’(x) f’(x)/f(x)

-2 1/9 0.1220681 1.10

-1 1/3 0.3662042 1.10

0 1 1.0986125 1.10

1 3 3.2958375 1.10

2 9 9.8875126 1.10

3 27 29.662538 1.10

Again, f’(x)/f(x) is constant, but this time it is equal to approx. 1.10 (greater than one).

Graphing f(x) and f’(x):• The derivative of f(x) = 3x is equal to 1.10 x 3x.

f(x) = 3x

f’(x) = 1.10 • 3x

As you can see on the graph, f’(x) is a vertical stretch of f(x) by a factor of 1.10.

Key Points:In general, for the exponential function f(x) = bx , we can conclude that… • f(x) and f’(x) are both exponential functions• the slope of the tangent at a point on the curve is

proportional to the value of the function at this point

• f’(x) is a vertical stretch or compression of f(x), dependent on the value of b

• the ratio f’(x)/f(x) is a constant and is equal to the stretch/compression factor

Determining the derivative of f(x)=bx:

Substitute into definition of the derivative

Product Rule (bx+h = bx • bh)

Common Factor

However, bx is constant as h0 and therefore does not

depend on h

Comparing y = bx and y = ex:• From 5.1 we know that the derivative of f(x) = ex was f’(x)= 1 • ex, therefore:

• From our first example, f(x) = 2x and f’(x) = 0.69 • 2x, therefore:

• Therefore, for f(x) = 3x and f’(x) = 1.10 • 3x:

Comparing y = bx and y = ex:• To determine a constant of proportionality we must

recall from 5.1:

• Consider ln 2 and ln 3:

These numbers match the constants that were determined in the two examples, by f’(x)/f(x)- therefore…

Derivative of f(x) = bx:

so

• Example 1: Determine the derivative ofa) f(x) = 5x

b) f(x) = 54x-5

a) Using the derivative of f(x) = bx:f’(x) = (ln 5) • 5x

b) f(x) = 54x-5 f(x) = 5g(x) g(x) = 4x-5 g’(x) = 4

Use chain rule and derivative of f(x) = bx.f’(x) = bg(x) • (ln b) • g’(x)f’(x) = 54x-5 • (ln 5) • 4f’(x) = 4(54x-5)ln 5

Derivative of f(x) = bg(x):

Yaaaaaaaaaaaaay! Almost done…

• Example 2: On January 1, 1950, the population of Swagtown was 50 000. The size of the population since then can be modelled by the function P(t) = 50 000(0.98)t, where t is the number of years since January 1, 1950.

a) What is the population of Swagtown on January 1, 2000?b) At what rate was the population of Swagtown changing on January 1,

2000? Was it increasing or decreasing?

a) January 1, 1950 January 1, 2000 = exactly 50 years, therefore t = 50.P(50) = 50 000(0.98)50

P(50) = 18 208.484

Therefore, the population on January 1, 2000 was around 18 208.

b) To determine the rate of change at a certain point in time, we require the derivative of P. Note: answer should be decreasing as this function has a base less than 1 (exponential decay)P’(t) = 50 000(0.98)t ln(0.98)P’(50) = 50 000(0.98)50 ln(0.98)P’(50) = -367.861

Therefore, after 50 years, the population was decreasing at a rate of approx. 368 people per year.