Section 3.5

1.Find the derivative of g (x) = x 2 ln x.

You will need to use the product rule.

f = x 2 f’ = 2x f is the first function.

s = ln x s’ = 1/x s is the second function.

f’ (x) = x 2 · 1/x + 2x ln x

= x + 2x ln x

If y f s, then y ' f s' s f '

2. Find the derivative of f(x) = ln x

f(x) = ln x = ln x ½ = ½ ln x.

You need to use the ln rule.

f’ (x) = ½ 1/x = 1/(2x)

1If y ln x, then y '

x

3. Find the derivative of f(x) = ln (- x)

You will need to use the general log rule.

Let u = - x then the chain is - 1

f’ (x) = (1/-x)· (- 1) = 1/x

1If y ln u, then y ' u'

u

4. Find the derivative of

You will need to use the general exponential rule.

x2x2exf )(

)()(' 2x2exf x2x2

u uIf y e , then y ' e u'

5. Find the derivative of f(x) = x – e – x.

You will need to use the general exponential rule on the second term.

f’(x) = 1 + e – x (-1) = 1 + e - x.

u uIf y e , then y ' e u'

6. Find the derivative of

You will need to use the general exponential rule.

xe1exf )(

x x1 e x 1 e xf '(x) e e e

u uIf y e , then y ' e u'

7. Find the derivative of f(x) = e 3.

e 3 is a constant so it derivative is 0.

8. Find the derivative of

You will need to use the product rule on the first term and the general exponential rule on the third term.

5ex2

1xxxf

2x22 ln)(

2x2 ex2x22

1xx2

x

1xxf ln)('

2xf '(x) 2x ln x 2xe

9. Find the derivative of f(x) = e x ln x 2.

You will need to use the product rule and the general ln rule on the second factor.

f’(x) = e x(2/x) + e x ln x 2 = 2e x/x + e x ln x 2

10. Find the derivative of f(x) = (e 2x + 1) 3.

You will need to use the general power rule and the general exponential rule.

f’(x) = 3 (e 2x + 1) 2 (2e2x) = 6e2x (e 2x + 1) 2

u uIf y e , then y ' e u' n nIf y u , then y ' nu u' and

11. Find the derivative of f(x) = (x 2 + 2 ln x) 1/2.

You will need to use the general power rule.

f’(x) = ½ (x 2 + 2 ln x) - ½ (2x + 2/x) = 2(x 2 + 2 ln x) - ½ (x + 1/x)

n nIf y u , then y ' nu u' and

12. Find the derivative of

Finally! you will need to use the quotient rule and the general exponential rule on the bottom term.

t = 10 t’ = 0

b = 1 + e – 2x b’ = -2e – 2x

x2e1

10xf

)(

2x2

x2

2x2

x2x2

e1

e20

e1

e2100e1xf

)()()())((

)('

2t (x) b(x)t '(x) b'(x)t (x)

If f (x) , then f '(x)b(x) b(x)

13. Given

a. Find f’(x)

Use the quotient rule

b. Find f’(1)

F’(1) = 1 from a calculator

5x

xxf

ln)(

10

44

25

45

x

xx5x

x

xx5x1xxf

lnln)/()(

2t (x) b(x)t '(x) b'(x)t (x)

If f (x) , then f '(x)b(x) b(x)

14. Given f(x) = ln (x 4 + 48)

a. Find f’(x)

Use the general ln rule.

b. Find f’(1)

F’(1) = 0.08163272 from a calculator

48x

x4x4

48x

1xf

4

33

4

)('

15. Given f(x) = ln (e x – 3x)

a. Find f’(x)

Use the general ln rule.

b. Find f’(0)

F’(0) = - 2 from a calculator

x3e

3e3e

x3e

1xf

x

xx

x

)('

16. Given f(x) = 5x ln x

a. Find f’(x)Use the product rule.

f’(x) = (5x)(1/x) + (5)(ln x) = 5 + 5 ln x

b. Find f’(2)

f’(2) = 8.4657357 from a calculator.

17. PERSONAL FINANCE: Earnings and Calculus. A recent study found that one’s earnings are affected by the mathematics courses one has taken. In particular, compared to someone making $40,000 who had taken no calculus, a comparable person who had taken x years of calculus would be earning $40,000 e 0.195x. Find the rate of change of this function at x = 1 and interpret your answer.

Graph the function on your calculator and find the derivative at 1.

f’ (1) = 9,479

A person with one year of calculus would earn $9,478 more per year if they had taken a second year of calculus.

18.PERSONAL FINANCE: Compound Interest (Sam Spartan’s IRA.). An investment of $5,000 at 10% interest compounded continuously will grow to A(t) = 5000 e 0.10t dollars in t years.

a. Find the rate of growth after 0 years.

b. Find the rate of growth after 20 years.

a. Graph the function on your calculator and find the derivative at 0.

f’ (0) = 500 It will grow $500 in the first year.

b. Graph the function on your calculator and find the derivative at 20.

f’ (20) = 3695 It will grow $3,695 in the 21st year.

19. GENERAL: Population. The world population (in billions) is predicted to be P(t) = 6.45e 0.0175t, where t is the number of years after 2005. find the instantaneous rate of change of the population in the year 2015 AND interpret your answer.

Graph the function on your calculator and find the derivative at 10.

f’ (10) = 0.134 the population will grow about .134 billion in the year after 2015.

20.BIOMEDICAL: Drug Dosage. A patient receives an injection of 1.2 milligrams of a drug, and the amount remaining in the bloodstream t hours late is A(t) = 1.2 e - 0.05t.

a. Find the instantaneous rate of change of this amount just after the injection (t = 0)

b. Find the instantaneous rate of change of this amount just after two hours.

Graph the function on your calculator and find the derivative at 0.

Graph the function on your calculator and find the derivative at 2.

The drug will be decreasing at .06 milligrams during the first hour.

The drug will be decreasing at .0543 milligrams during the third hour.

21. BUSINESS: Sales. The weekly sales (in thousands) of a popular product are predicted to be S(x) = 1000 - 900 e - 0.1x after x weeks.

a. Find the rate of change of sales after 1 week AND interpret your answer.

b. Find the rate of change of sales after 10 weeks.AND interpret your answer.

Graph the function on your calculator and find the derivative at 1.

Graph the function on your calculator and find the derivative at 10.

The sales will be increasing at a rate of 81 during the second week.

The sales will be increasing at a rate of 33 during the eleventh week.

22. ECONOMICS: Consumer Expenditure. If consumer demand for a product is given by D(p) = 5000 e -0.01p, where p is the selling price in dollars, find the price that maximizes consumer expenditure.

You need to read carefully. You are asked to maximize consumer expenditure. Remember that expenditure is the price times the demand. So you want to maximize

E(p) = p 5000 e -0.01p.

Use your calculator.

The price that maximizes the expenditure is $100.

23.BUSINESS: Maximum Revenue. The price function for a company is P(x) = 400e - 0.20x, where p is the price (in dollars) where x (in thousands) is the number sold.

a. Find the revenue function.

b. find the quantity and the price that will maximize the revenue.

a. R(x) = x p = 400xe - 0.20x

Graph the revenue function on your calculator and find the maximum.

The revenue will be a maximum of $735.76 when x = 5.

Substituting 5 into the given price equation will yield a price of

P(5) = 400e (- 0.20)(5) = $147.15