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

Section 5.2: Derivative of the General Exponential Function, y = b x. Dana, Dina, Isabella. Consider the function f(x) = 2 x. 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):.

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

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  1. Section 5.2: Derivative of the General Exponential Function, y = bx Dana, Dina, Isabella

  2. Consider the function f(x) = 2x. 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?

  3. 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.

  4. Consider the function f(x) = 3x. Again, f’(x)/f(x) is constant, but this time it is equal to approx. 1.10 (greater than one).

  5. 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.

  6. 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

  7. 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 h0 and therefore does not depend on h

  8. 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:

  9. 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…

  10. Derivative of f(x) = bx: so

  11. Example 1: Determine the derivative of • f(x) = 5x • 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) • 4 f’(x) = 4(54x-5)ln 5

  12. Derivative of f(x) = bg(x): Yaaaaaaaaaaaaay! Almost done…

  13. 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. • What is the population of Swagtown on January 1, 2000? • 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)tln(0.98) P’(50) = 50 000(0.98)50ln(0.98) P’(50) = -367.861 Therefore, after 50 years, the population was decreasing at a rate of approx. 368 people per year.

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