2.3 The Chain Rule and Non

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Section 2.3
1. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) = (x 2 + 1) 3
f ( x) = (x2 +1)3
f ′( x) = 3(x2 + 1)2(2x) = 6x(x2 + 1)2
2. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) = (3x 2 – 5x + 2) 4
f (x) = (3x2 − 5x + 2)4
f ′( x) = 4(3x2 − 5x + 2)3 (6x − 5)
3. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) 
f (x) 
3
f '(x) 
3
9x  1
9 x  1  (9 x  1)
1
3
(9 x  1)
 2
3
1
3
(9 )  3 (9 x  1)
 2
3
4. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) = (4 - x 2 ) 4
y = (4 − x2 )4
y ′ = 4(4 − x2 )3(−2x) = −8x(4 − x2)3
5. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) = x 4 + (1 – x ) 4
y = x 4 + (1 − x)4
y ′ = 4x3 + 4(1 − x)3(−1) = 4x 3 − 4(1− x)3
6. Use the Generalized Power Rule – Chain Rule – to find the derivative of
1
f (x) 
3
1
f (x) 
3
f '(x) 
(9 x  1)
2
3
(9 x  1)
 (9 x  1)
2
2 3
2
(9 x  1)
5 3
(9 )   6 (9 x  1)
5 3
7. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) = [ (x 2 + 1 ) 3 + x] 3
f ( x) = [(x2 +1)3 + x]3
f ′( x) = 3[( x2 + 1)3 + x]2  dy/dx[(x2 + 1)3 +x]
= 3[( x2 + 1)3 + x]2 [3(x2 + 1)2 (2x) +1]
= 3[( x2 + 1)3 + x]2 [6x(x2 + 1)2 + 1]
8. Use the Generalized Power Rule – Chain Rule – to find the derivative of
f (x) = (2x + 1) 3 (2x – 1) 4
Product Rule and Chain Rule combined.
f ( x) = (2x + 1)3(2x − 1)4
f ′( x) = (2x + 1)3[4(2x − 1)3(2)] + (2x − 1)4 3(2x + 1)2 (2)
f ( x) = 8(2x + 1)3(2x −1)3 + 6(2x + 1)2(2x −1)4
9. Use the Generalized Power Rule – Chain Rule – to find the derivative of
 x  1
f (x)  

x

1


3
2
 x  1  ( x  1)(1)  ( x  1) (1)
f '(x)  3 

2
( x  1)
 x  1
Quotient Rule
Chain Rule
2
2
 x  1
2
 6 ( x  1)
f '(x)  3 


2
2
( x  1)
 x  1  ( x  1)
You could also do this problem as a product rule by rewriting it as:
f (x )  (x  1)  (x  1)
3
3
10. BUSINESS: Cost - A company’s cost function is given below in dollars, where x
is the number of units. Find the marginal cost function and evaluate it at 20.
C (x) 
4 x  900
2
The marginal cost function is the derivative of the function C (x)
C (x )  (4x  900)
2
C '(x) 
1
1 2
(4 x  900 )
2
1 2
(8 x )  4 x (4 x  900 )
2
1 2
2
C ' (20)  4 (20)[4(20)  900]
2
1 2
 1.6
Since you are trying to calculate a
derivative at a point you may use your
calculator. Graph the original cost
function and go to the derivative menu.
Do you know what the 1.6 means????
11. SOCIOLOGY: Income Status – A study estimated how a person’s social status
(rated on a scale where 100 indicates the status of a college graduate) depends
upon income. Based on this study, with a an income of x thousand dollars, a
person’s status is given by S (x) = 17.5 (x – 1) 0.53. Find S ’ (25) and interpret your
answer.
S(x) = 17.5(x −1)0.53
S ′ (x) = 9.275(x −1) −0.47
S ′ (25) = 9.275(25 − 1) −0.47 ≈ 2.08
At an income of $25,000 social status increases
by about 2.08 units per additional $1000 of
income.
Since you are trying to calculate a
derivative at a point you may use your
calculator. Graph the original cost
function and go to the derivative menu.
This is a graph of the
original equation S(x).
12. BIOMEDICAL: Drug Sensitivity – The strength of a patient’s reaction to a dose
of x milligrams of a certain drug is R (x) = 4x (11 + 0.5 x) 0.5 for 0 ≤ x ≤ 140. The
derivative R‘ (x) is called the sensitivity to the drug. Find R‘ (50), the sensitivity
to a dose of 50 milligrams of the drug.
R ( x )  4 x (11  0 . 5 x )
0 .5
You need to use the product rule!
R ' (x )  (4x ) 0.5 (11  0.5 x )
 x (11  0.5 x )
R (50 )  50 (11  25 )
 0.5
 0.5
 0.5
(0.5)  4 (11  0.5 x )
 4 (11  0.5 x )
 4 (11  25 )
0.5
0.5

50
6
Since you are trying to calculate a
derivative at a point you may use your
calculator. Graph the original cost
function and go to the derivative menu.
0.5
 4 6  32.33
13. ENVIRONMENTAL SCIENCE: Pollution – The carbon monoxide level in a city
is predicted to be 0.02 x 2/3 + 1 ppm (parts per million), where x is the population
in thousands. In t years the population of the city is predicted to be x (t) 12 + 2t
thousand people. Therefore, in t years the carbon monoxide level will be
P 9t) = 0.02 (12 + 2t) 3/2 + 1 ppm
Find P ‘ (2), the rate at which carbon monoxide pollution will be increasing in 2
years.
P(t ) = 0.02(12 + 2t)3/2 + 1
P′ (t ) = 0.03(12 + 2t ) 1/2 (2) = 0.06(12 + 2t) 1/2
P′ (2) = 0.06[12 + 2(2)]1/2 = 0.24
Since you are trying to calculate a
derivative at a point you may use your
calculator. Graph the original cost
function and go to the derivative menu.
14. For the following function below, find the x-values at which the
derivative does not exist.
The derivative does not exist at the corner points, end
points or gaps: x = − 3, 1, 4.
15. Why is the following function not differentiable at x = 0?
f (x) = x 2/3.
Graph the function on your calculator.
There is a corner point at x = 0.
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