Hints and Solutions to Selected Problems



Hints and Solutions to Selected Problems

Derivative Techniques – 1

5. Note that g(x) = 125x3 6. Note that y = 3t2 – 2t(2 7. Note that f(u) = u3/2 + 3u1/2 + 4u–1/2

8. Need a point and a slope:

Point: When x = 2, y = 6. Slope: y = x + 8x–1 ( y' = 1 – 8x–2; when x = 2, y' = –1.

Tangent line equation: y – 6 = –(x – 2)

9. Need a point and a slope:

Point: When x = 1, y = 2. Slope: y = x2 + x1/2 ( y' = 2x + 0.5x–1/2 ; when x = 1, y' = 2.5.

Tangent line equation: y – 2 = 2.5(x – 1)

10a. Remember, velocity is the rate of change (derivative) of position.

b. And acceleration is the rate of change of velocity (second derivative of position).

c. v = 3t2 – 3 = 0 when t = 1. a(1) = 6 m/s2

11. [pic]

Multiply by g(x) to clear fraction: [pic]

Solve for q': [pic]

[pic]

This is usually written: [pic]

Derivative Techniques – 2

4 – 8. Do not use the product and quotient rules for problems of his type. The rules do work but they are unnecessary, make the problems harder and lead to mistakes. If either term of a product is a constant, you do not need the product rule. If the denominator of a quotient is a constant, you do not need the quotient rule. If the numerator of a quotient is a constant, you do not need the quotient rule if the denominator is a power of x and you won’t need it for any denominators (with constant numerator) after we learn the chain rule. DON’T MAKE PROBLEMS HARDER THAN THEY NEED TO BE!

4. 3 is a constant; do not use the product rule. [pic] (which you should be able to do by now in one step in your head).

5. 2 is a constant; do not use the quotient rule.[pic] (which, again, you should be able to do in one step in your head).

6. 12 (or (12) is a constant; do not use the quotient rule. [pic] (This one maybe two steps: rewrite in exponent form, then take derivative.)

7. a and b are constants, do not use the product rule.

[pic]

8. a and b are constants, do not use the quotient rule.

[pic]

9. Need 1) A point. At x = (2, [pic].

2) The slope. For now, because the denominator is a binomial, we need the quotient rule. We will soon learn a better way.[pic]. [pic]

Tangent line: [pic]

10a. [pic] means: Differentiate [pic]and evaluate the result for x = 4.

11a. [pic]

b. [pic]

12a. [pic]

b. [pic]

Derivative Techniques – 4

8. Remember: [pic], [pic] and [pic]. YOU SHOULD KNOW THIS.

9. Remember: cos( = –1 and sin( = 0

Derivative Techniques – 5

1. a. v(t) = x'(t) = 3t2 – 24t + 36

b. “At rest” means v = 0. v = 3(t – 2)(t – 6) = 0 at time t = 2 and t = 6.

c. “Forward” means v > 0. Bug is moving forward for 0 < t < 2

and t > 6.

d. a(t) = v'(t) = 6t – 24

4a. The easiest way is to use nDeriv on your calculator.

c. [pic] ( [pic]. Since this negative and increasing (until the water is gone), water flows out the fastest (240 gal/min) at t = 0 and the slowest

(0 gal/min) at t = 40 min.

5c. [pic]

6. a. [pic]. [pic] Pa.

b. [pic]. The volume is decreasing; as pressure increases, the volume decreases more slowly. (Mathematically, the rate of change of the volume is increasing (becoming “less negative.”))

Derivative Techniques – 6

1a. [pic] means the derivative of 3f(x) – 2g(x) evaluated at x = 1.

e. Using a symmetric difference quotient: [pic]

2a. (fg)'(4) = f(4)g'(4) + g(4)f '(4) = (2.5)(1/3) + (2/3)(–1/2) = 1/2

b. [pic]

4c. Use a tangent line approximation: y – f(4) = f '(4)(r – 4) ( y – 268 = 201(r – 4)

Derivative Techniques – 7

1. Mommy = ( )4; Baby = (5x + 1) 2. Mommy = 3e( ); Baby = (2t

3. Mommy = 2sin ( ); Baby = 3( 4. Mommy = ln ( ); Baby = ex ( 1

5. Mommy = 4( )(2; Baby = 1 + 3r 6. Mommy = [pic]; Baby = x2 + 4

7. Mommy = (ln( ); Baby = cos( 8. Mommy = ( )2; Baby = x2 ( 3x + 5

9. Mommy = 6tan( ); Baby = [pic] 10. Mommy = 4e( ); Baby = [pic]

11. Mommy = [pic]; Baby = u2 – 2u + 5 12. Mommy = sec ( ); Baby = 5x2

13. Rewrite as [pic]. Then Mommy = 4( )3; Baby = cosx

14. Mommy = ( )(1/3; Baby = 4r + 1

15. You need to use both the product rule and the chain rule.

For the derivative of [pic], Mommy = e( ); Baby = –z2.

16. You need to use the quotient rule and the chain rules.

For numerator, Mommy = sin( ) and Baby = [pic]; for denominator, Mommy = e( ) + 1, Baby = x2

17. If [pic] so [pic]

18a.[pic]

b. [pic]

c. [pic]

d. [pic] but g'(2) is undefined.

e. [pic]

Derivative Techniques – 8

5. Rewrite as [pic]. Then Mommy = 12( )2, Baby = tan( ) and Baby’s baby = [pic]

[pic]

6. Rewrite as [pic]. Then Mommy = [pic], Baby = 1 + ( )2, Baby’s baby = sinx

[pic]

7. y = ln(sec x + tan x) ( [pic]

12. [pic] ( [pic]

Derivative Techniques – 9

1. 3yy' + y' = 3 2. Remember the product rule: 2x – (xy' + y) + 3yy'= 0

3. Don’t look here. Ask Jess or Chris.

4. Where y = 0, we get [pic] ( x = 2. Differentiating (and using the product rule twice), [pic]. At (2, 0), [pic] ( [pic]

5. [pic]. At [pic], [pic] (

[pic] ( [pic]

6. When x = 2, [pic] ( y4 ( 12y ( 117 = 0 ( y = (3 (we want quadrant IV).

[pic]. At (2, (3), [pic] ( [pic].

Tangent line: [pic].

7. [pic] ( [pic]. For horizontal tangents, we need y' = 0 ( x + 3y = 0.

We need points on the ellipse where x = (3y. Subbing into the original equation,

[pic] ( 16y2 = 16 ( y = (1. Then x = (3y = [pic]. The horizontal tangents are at [pic]. (If you couldn’t finish this problem without looking at the solution, here is a bonus for you: Find the points where the ellipse has vertical tangents.)

Derivative Techniques – 11

1. [pic] so [pic] so choice B is correct. (You can check that the other choices don’t work. Be careful of choice E, it says 0 = 1/2.)

2. [pic]. We need to know f –1(3): x3 + x – 7 = 3 ( x = 2. Now we need f '(2) = 13. So

[pic]

3. [pic]. Need f –1(15): [pic] ( x = 4. f '(4) = 8.25

so g’(15) = 1/8.25 = 4/33

4. Differentiating implicitly: 1 = 3yy' + 2y'. When x = 7, y = 1 so y'(7) = 1/5.

5. [pic]. f –1(3) = 3.1 and [pic] so [pic]

6. [pic]. To find f –1(3), go to y = 3 on the graph of f: x = 1. f '(1) = –1/2

so (f –1)'(3) = –2.

7a. If y = sin–1x, then x = sin y, 1 = (cos y) y' and y' = sec y = [pic].

b. If y = ln x, then x = ey, 1 = eyy' and y' = [pic].

Derivative Techniques – 13

1. First note that f is continuous at x = 1: f(1) is defined, [pic] exists and [pic].

Now, [pic]. Since [pic] and [pic], f has a “corner” at x = 1.

2. We need f to be continuous so [pic] must exist: [pic] ( 3(1)2 = a(1) + b.

We also need [pic] ( 6(1) = a. This gives a = 6, b = –3.

3. Note that f is continuous at x = 0. Now [pic] and [pic] so f '(0) = 0.

But [pic], [pic] and [pic] so f ' has a “corner” at x = 0 and [pic] is undefined.

Derivative Techniques – Review

1c. [pic]

f. [pic]

g. [pic]

5c. [pic]

[pic]

d. [pic]

[pic]

f. [pic]

6. [pic]

7. When y = e2, [pic] ( 2x = 2e + 4e ( x = 3e. Differentiating: [pic] (

[pic] ( [pic] ( [pic] ( y' = (2e. Tangent line: y ( e2 = (2e(x ( 3e)

9a. First, when x = –4, –64 + 12y + y2 +16 = 0 ( y = 3.

Differentiating implicitly: 3x2 – 3(xy' + y) + 2yy' = 0. At (–4, 3): 48 – 3(–4y' + 3) + 6y' = 0 ( y' = –13/6

So the tangent line equation is [pic]

11. For continuity at x = 1, we need [pic] or [pic].

The derivative is [pic]. For differentiability at x = 1, we need [pic] or [pic] which gives [pic].

13. Differentiating, we get [pic] ( [pic].

For a horizontal tangent, we need y' = 0 ( 2x ( 4 = 0 ( x = 2. When x = 2, 2y3 ( 6y + 5 = 0 ( y = (2.054 so the only horizontal tangent is at (2, (2.054).

For vertical tangents, we need y' undefined ( 6y2 ( 6 = 0 ( y = (1. When y = 1, (x2 + 4x ( 3 = 0 ( x = 1 or x = 3. When y = (1, (x2 + 4x + 5 = 0 ( x = (1 or x = 5. So there are four horizontal tangents: (1, 1), (3, 1), ((1, (1) and (5, (1). (Note: We should really sub all five points back into the derivative formula to make sure none of them make the derivative an indeterminate form, [pic]. If y' is indeterminate, we cannot draw any conclusions about our tangent lines at that point. None of these points make y' indeterminate.)

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0 2 6

x

y

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1

[pic]

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