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1. 2.

4.

6.

7.8.

9.10.

11. 12.

13. 14.

15.

17. 18.

19.

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2(𝑑𝑑 + 1)(2𝑑𝑑2 + 𝑑𝑑 + 1) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=𝑑𝑑 + 12√𝑑𝑑

+ √𝑑𝑑 + 2

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 3𝑑𝑑2 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= (𝑑𝑑7 + 15)2(22𝑑𝑑7 + 15)

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 𝑑𝑑(𝑑𝑑 + 7)2(5𝑑𝑑 + 14) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=32𝑑𝑑8 + 49𝑑𝑑6

√4𝑑𝑑2 + 7

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=3𝑑𝑑2 + 4

2√𝑑𝑑

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=2𝑑𝑑3

√𝑑𝑑4 − 1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 2(2𝑑𝑑 − 1)3(10𝑑𝑑 + 3) 𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

= 1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=1

(𝑑𝑑 + 1)2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=3𝑑𝑑2 − 2𝑑𝑑(3𝑑𝑑 − 1)2

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=20𝑑𝑑4 + 59𝑑𝑑2 − 14

(5𝑑𝑑2 + 2)2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=4𝑑𝑑(𝑑𝑑2 − 1)2(𝑑𝑑2 + 2)

(𝑑𝑑2 + 1)2

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=9𝑑𝑑8√𝑑𝑑2 − 1 − 𝑑𝑑(𝑑𝑑9

2− 1)

√𝑑𝑑 − 1𝑑𝑑2 − 1

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=−4𝑑𝑑4 + 24𝑑𝑑

(𝑑𝑑3 + 3)2

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=52

52𝑑𝑑 + 3𝑑𝑑2

(√𝑑𝑑 + 1)2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=−2

(𝑑𝑑 − 1)2

𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=2𝑑𝑑

(𝑑𝑑2 + 4)2𝑑𝑑𝑑𝑑𝑑𝑑𝑑𝑑

=

12− √𝑑𝑑 + 7

�𝑑𝑑 + 72�

24√𝑑𝑑

Differentiation Using Product and Quotient Rule 1

.

21. Find the horizontal tangents of: 𝑦 = 𝑥4 − 2𝑥2 + 2

Horizontal tangents occur when slope = zero.

𝑦′ = 4𝑥3 − 4𝑥 = 0

𝑥(𝑥 + 1)(𝑥 − 1) = 0

𝑥 = 0, −1, 1

𝑃𝑜𝑖𝑛𝑡𝑠: (0,2) (−1,1) (1,1)

(The function is even, so we only get two horizontal tangents.)

22. Find equations of the tangent line and normal line to the curve y = √𝑥

1+𝑥2 at the point (1, ½).

slope of the tangent line at (1, ½) :

tangent line at (1, ½):

𝑦 − 1 = −1

4(𝑥 − 1) → 𝑦 = −

1

4𝑥 +

3

4

normal line at (1, ½):

𝑦 −1

2= 4(𝑥 − 1) → 𝑦 = 4𝑥 −

7

2

23. At what points on the hyperbola xy = 12 is the tangent line parallel to the line 3x + y = 0?

Since xy = 12 can be written as y = 12/x, we have:

𝑑𝑦

𝑑𝑥= 12(𝑥−1)′ = −

12

𝑥2

Let the x-coordinate of one of the points in question be 𝑎.

Slope of the tangent line at that point is −12

𝑎2 , and that has to be equal to the slope of line

3x + y = 0

−12

𝑎2 = −3 𝑜𝑟 𝑎 = ±2

the required points are: (2, 6) and (-2, -6)

1. State the Product Rule: d f x g xdx

2. State the Quotient Rule: f xd

dx g x

3. Find the derivative of each.

(a) 36 5 3f x x x (b) 22 sin cosh t t t t t

(c) 22 cotf x x x

(d) tansin 1x xf xx

(e) 2

22

3 11

x xf x x xx

(f) tan sinf x x x

__ __ _____

𝑑

𝑑𝑥[𝑓(𝑥)𝑔(𝑥)] = 𝑓′(𝑥)𝑔(𝑥) + 𝑓(𝑥)𝑔′(𝑥)

𝑑

𝑑𝑥[𝑓(𝑥)

𝑔(𝑥)] =

𝑓′(𝑥)𝑔(𝑥) − 𝑓(𝑥)𝑔′(𝑥)

𝑔2(𝑥)

𝑓′(𝑥) = 6(𝑥3 − 3) + (6𝑥 + 5)(3𝑥2) ℎ′(𝑥) = 2 sin 𝑡 + 2 𝑡 cos 𝑡 + 2𝑡 cos 𝑡 − 𝑡2 sin 𝑡

𝑓′(𝑥) = 4𝑥 cot 𝑥 − 2𝑥2 𝑐𝑠𝑐2𝑥

𝑓′(𝑥) =(sin 𝑥 + 1)(1 + 𝑠𝑒𝑐2𝑥) − (𝑥 + tan 𝑥)(cos 𝑥)

=

(sin 𝑥 + 1)2

sin 𝑥 ∙ 𝑠𝑒𝑐2𝑥 + 1 + 𝑠𝑒𝑐2𝑥 − 𝑥 𝑐𝑜𝑠 𝑥 (sin 𝑥 + 1)2

(2𝑥 − 1)(𝑥2 + 1) − (𝑥2 − 𝑥 − 3)(2𝑥)𝑓′(𝑥) = [

(𝑥2 + 1)2

𝑥2 − 𝑥 − 3] [𝑥2 + 𝑥 + 1] + [

𝑥2 + 1] [2𝑥 + 1]

𝑓′(𝑥) = 𝑠𝑒𝑐2𝑥 ∙ sin 𝑥 + tan 𝑥 ∙ cos 𝑥

= sec 𝑥 ∙ tan 𝑥 + sin 𝑥

= 24𝑥3 + 15𝑥2 − 18 = 2 sin 𝑡 + 4 𝑡 cos 𝑡 − 𝑡2 sin 𝑡

Product & Quotient Rules with Trig

5. Find the equation of the tangent line to f x x 1x2 1 at the point where f x crosses the x-axis.

(g) f x 2cos____xx

(h) h x csc2 x

4. Evaluate f

4 if f x sin x sin x cos x , then find the equation of the tangent line at

4x .

𝑓′(𝑥) =(− sin 𝑥)(𝑥2) − (cos 𝑥)(2𝑥)

(𝑥2)2

=−𝑥2 sin 𝑥 − 2𝑥 cos 𝑥

𝑥4

= (0 − 1 ∙ cos 𝑥

𝑠𝑖𝑛2𝑥) (

1

sin 𝑥) + (

1

sin 𝑥) (

0 − 1 ∙ cos 𝑥

𝑠𝑖𝑛2𝑥)

= −2cot 𝑥

𝑠𝑖𝑛2𝑥

ℎ′(𝑥) =𝑑

𝑑𝑥[(

1

sin 𝑥) (

1

sin 𝑥)]

𝑃 (𝜋

, 1) 4

𝑓′(𝑥) = cos 𝑥(sin 𝑥 + cos 𝑥) + sin 𝑥(cos 𝑥 − sin 𝑥)

= 2 sin 𝑥 cos 𝑥 + 𝑐𝑜𝑠2𝑥 − 𝑠𝑖𝑛2𝑥

= 2 (√2

2) (

√2

2) + (

√2

2)

2

− (√2

2)

2

= 1

𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑎𝑡 𝑃 (𝜋

4, 1) : 𝑦 = 𝑥 + (1 −

𝜋

4)

𝑥 − 𝑖𝑛𝑡𝑒𝑟𝑠𝑒𝑐𝑡𝑖𝑜𝑛: (𝑥 − 1)(𝑥2 + 1) = 0 ∴ 𝑃(1,0)

𝑓′(𝑥) = (1)(𝑥2 + 1) + (𝑥 − 1)(2𝑥) ∴ 𝑓′(1) = 2

𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑎𝑡 𝑃(1,0): 𝑦 = 2𝑥 − 2

6. 11

y xx

that are parallel to the line 2y x 6 .

7. If 3x2

f xx

and g x 5x 4x 2

, verify that f x gx , and explain the relationship between f

and g.

Find the equation of the tangent lines to the graph of

𝑎𝑙𝑙 𝑝𝑎𝑟𝑎𝑙𝑙𝑒𝑙 𝑙𝑖𝑛𝑒𝑠 ℎ𝑎𝑣𝑒 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑠𝑙𝑜𝑝𝑒

𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑡𝑎𝑛𝑔𝑒𝑡 𝑙𝑖𝑛𝑒 𝑖𝑠 −1

2

𝑦′ =(1)(𝑥 − 1) − (𝑥 + 1)(1)

(𝑥 − 1)2

=−2

(𝑥 − 1)2

∴ −2

(𝑥 − 1)2= −

1

2 → (𝑥 − 1)2 = 4 → (𝑥 − 1) = ± 2 → 𝑥 = −1 & 𝑥 = 3

𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑎𝑡 (−1,0) 𝑜𝑟 (3,2): 𝑦 = −1

2𝑥 −

1

2 & 𝑦 = −

1

2𝑥 +

7

2

𝑓′(𝑥) =(3)(𝑥 + 2) − (3𝑥)(1)

(𝑥 + 2)2

=6

(𝑥 + 2)2

𝑔′(𝑥) =(5)(𝑥 + 2) − (5𝑥 + 4)(1)

(𝑥 + 2)2

=6

(𝑥 + 2)2

𝑓(𝑥) =𝑥 + 2

=𝑥 + 2

=3𝑥 3(𝑥 + 2) − 6 −6

𝑥 + 2+ 3

𝑔(𝑥) =5𝑥 + 4

𝑥 + 2=

5(𝑥 + 2) − 6

𝑥 + 2=

−6

𝑥 + 2+ 5

𝐼𝑓 𝑓′(𝑥) = 𝑔′(𝑥), 𝑡ℎ𝑒𝑛 𝑓(𝑥)𝑎𝑛𝑑 𝑔(𝑥)𝑜𝑛𝑙𝑦 𝑑𝑖𝑓𝑓𝑒𝑟 𝑏𝑦 𝑎 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡. 𝑃𝑟𝑜𝑜𝑓:

____

8. Determine whether there exist any values of x in the interval 0,2 such that the rate of change of

secf x x and the rate of change of cscg x x are equal.

9. Sketch the graph of a differentiable function f such that 2 0f , 0f for 2x , and 0f for2x

10. If 2 3g , 2 2g , 2 1h , and 2 4h , find 2f for

(a) 2f x g x h x (b) 4f x h x (c) g x

f xh x (d) 2f x g x h x

𝑓′(𝑥) =𝑑

𝑑𝑥[

1

cos 𝑥] =

(0)(cos 𝑥) − (1)(− sin 𝑥)=

sin 𝑥

𝑐𝑜𝑠2𝑥

𝑔′(𝑥) =𝑑

𝑑𝑥[

1

sin 𝑥] =

𝑐𝑜𝑠2𝑥

(0)(sin 𝑥) − (1)(cos 𝑥)

𝑠𝑖𝑛2𝑥= −

cos 𝑥

𝑠𝑖𝑛2𝑥

𝑠𝑖𝑛 𝑥= −

cos 𝑥

𝑐𝑜𝑠2𝑥 𝑠𝑖𝑛2𝑥

𝑠𝑖𝑛3𝑥 = −𝑐𝑜𝑠3𝑥 → sin 𝑥 = − cos 𝑥

𝑥 =3𝜋

4 & 𝑥 =

7𝜋

4

𝑎𝑛𝑠𝑤𝑒𝑟𝑠 𝑤𝑖𝑙𝑙 𝑣𝑎𝑟𝑦

𝑓′(2) = 0 𝑓′(2) = −4 𝑓′(2) = −10 𝑓′(2) = 28

____

𝑝𝑜𝑖𝑛𝑡 𝑜𝑓 𝑖𝑛𝑡𝑒𝑟𝑐𝑒𝑝𝑡: 𝑎 + 𝑏 + 𝑐 = 0

𝑔𝑟𝑎𝑝ℎ 𝑝𝑎𝑠𝑠𝑒𝑠 𝑡ℎ𝑟𝑜𝑢𝑔ℎ 𝑡ℎ𝑒 𝑝𝑜𝑖𝑛𝑡 (2,7): 4𝑎 + 2𝑏 + 𝑐 = 7

𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑎𝑡 (2,7) ℎ𝑎𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 10: 𝑓′(𝑥) = 2𝑎𝑥 + 𝑏 → 4𝑎 + 𝑏 = 10

𝑎 + 𝑏 + 𝑐 = 04𝑎 + 2𝑏 + 𝑐 = 74𝑎 + 𝑏 = 10

𝑎 = 3 𝑏 = −2 𝑐 = −1

𝑓(𝑥) = 3𝑥2 − 2𝑥 − 1

11. Find a second-degree polynomial f x ax2 bx c such that f x has an x-intercept at x 1 , and

the graph of f x has a tangent line with a slope of 10 at the point 2,7

12. If the normal line to the graph of a function f at the point 1,2 passes through the point 1,1 , then

what is the value of f 1 ?

𝑓′(1) 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑙𝑜𝑝𝑒 𝑜𝑓 𝑡𝑎𝑛𝑔𝑒𝑛𝑡 𝑙𝑖𝑛𝑒 𝑎𝑡 𝑥 = 1. 𝐼𝑡 𝑖𝑠 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑎𝑙 𝑜𝑓 𝑡ℎ𝑒 𝑛𝑜𝑟𝑚𝑎𝑙 𝑙𝑖𝑛𝑒 𝑠𝑙𝑜𝑝𝑒:

𝑚𝑡 = −1

𝑚𝑛

𝑚𝑛 =2 − 1

=1

21 − (−1)

𝑓′(1) = 𝑚𝑡 = −2

13. If y x sin x , find3

3d ydx

𝑑3𝑦

𝑑𝑥3= −3 sin 𝑥 − 𝑥 cos 𝑥

𝑦 = cos 𝑥

𝑦′ = −𝑠𝑖𝑛 𝑥

𝑦′′ = − cos 𝑥

𝑦′′′ = 𝑠𝑖𝑛 𝑥

𝑦(4) = cos 𝑥 = 𝑦

999 = 249 ∙ 4 + 3

𝑦(999) = 𝑦′′′ = 𝑠𝑖𝑛 𝑥

𝑦 = 𝑠𝑖𝑛 𝑥

𝑦′ = cos 𝑥

𝑦′′ = − 𝑠𝑖𝑛 𝑥

𝑦′′′ = −𝑐𝑜𝑠 𝑥

𝑦(4) = 𝑠𝑖𝑛 𝑥 = 𝑦

725 = 181 ∙ 4 + 1

14. Find the following

(a) x999

999d cosdx

(c) x725

725d sindx

𝑦(999) = 𝑦′ = 𝑐𝑜𝑠 𝑥

15.

dx

If sin 2x 2sin x cos x and cos 2x cos2 x sin2 x ( (re)memorize these), use these identities toevaluate the following using the product rule.

(a) d sin 2x (b) d cos 2xdx

𝑑𝑥[𝑠𝑖𝑛 2𝑥] =

𝑑 𝑑[2 sin 𝑥 cos 𝑥]

𝑑𝑥

= 2(cos 𝑥)(cos 𝑥) + 2(sin 𝑥)(− sin 𝑥)

= 2(𝑐𝑜𝑠2𝑥 − 𝑠𝑖𝑛2𝑥)

= 2 cos 2𝑥

𝑑

𝑑𝑥[cos 2𝑥] =

𝑑

𝑑𝑥[𝑐𝑜𝑠2𝑥 − 𝑠𝑖𝑛2𝑥]

=𝑑

[(𝑐𝑜𝑠 𝑥)(𝑐𝑜𝑠 𝑥) − (𝑠𝑖𝑛 𝑥)(𝑠𝑖𝑛 𝑥)] 𝑑𝑥

= [(−𝑠𝑖𝑛 𝑥)(𝑐𝑜𝑠 𝑥) + (𝑐𝑜𝑠 𝑥)(− 𝑠𝑖𝑛 𝑥)] − [(𝑐𝑜𝑠 𝑥)(𝑠𝑖𝑛 𝑥) + (𝑠𝑖𝑛 𝑥)(𝑐𝑜𝑠 𝑥)]

= −2 sin 𝑥 cos 𝑥 − 2 sin 𝑥 cos 𝑥 = − sin 2𝑥 − sin 2𝑥

= −2 sin 2𝑥

𝑏. 𝑑

𝑑𝑥[cos 2𝑥]

2

SOLUTIONS

1. lim ℎ→0

(𝑥𝑥 + ℎ)2 − 𝑥𝑥2

ℎ=

𝑑𝑑𝑑𝑑𝑥𝑥

𝑥𝑥2 = 2𝑥𝑥

2. lim ℎ→0

tan (𝑥𝑥 + ℎ) − tanℎℎ

=𝑑𝑑𝑑𝑑𝑥𝑥

tan 𝑥𝑥 =1

𝑠𝑠𝑠𝑠𝑠𝑠2 𝑥𝑥

3. lim ℎ→0

(𝑥𝑥 + ℎ)4 − 𝑥𝑥4

ℎ=

𝑑𝑑𝑑𝑑𝑥𝑥

𝑥𝑥4 = 4𝑥𝑥3

4. lim ℎ→0

2(𝑥𝑥 + ℎ)2 + 3(𝑥𝑥 + ℎ) − 2𝑥𝑥2 − 3𝑥𝑥ℎ

=𝑑𝑑𝑑𝑑𝑥𝑥

(2𝑥𝑥2 + 3𝑥𝑥) = 4𝑥𝑥 + 3

5. lim ℎ→0

sin[2(𝑥𝑥 + ℎ)]− sin 2𝑥𝑥ℎ

=𝑑𝑑𝑑𝑑𝑥𝑥

sin 2𝑥𝑥 = 2 cos 2𝑥𝑥

6. lim ℎ→0

cos(𝑥𝑥 + ℎ)2 − cos 𝑥𝑥2

ℎ=

𝑑𝑑𝑑𝑑𝑥𝑥

cos 𝑥𝑥2 = −2𝑥𝑥 sin 𝑥𝑥2

7. lim 𝑥𝑥→3𝑥𝑥2−9𝑥𝑥−3

= lim 𝑥𝑥→3𝑥𝑥2−32

𝑥𝑥−3= � 𝑑𝑑

𝑑𝑑𝑥𝑥 𝑥𝑥2�

𝑥𝑥=3= 6

8. lim 𝑥𝑥→5

𝑥𝑥2 − 52

𝑥𝑥 − 5= �

𝑑𝑑𝑑𝑑𝑥𝑥

𝑥𝑥2�𝑥𝑥=5

= 10

9. lim 𝑥𝑥→5

𝑥𝑥2 − 25𝑥𝑥 − 5

= lim 𝑥𝑥→5

𝑥𝑥2 − 52

𝑥𝑥 − 5= �

𝑑𝑑𝑑𝑑𝑥𝑥

𝑥𝑥2�𝑥𝑥=5

= 10

10. lim 𝑥𝑥→𝜋𝜋/3

sin 𝑥𝑥 − sin𝜋𝜋3𝑥𝑥 − 𝜋𝜋

3= �

𝑑𝑑𝑑𝑑𝑥𝑥

sin 𝑥𝑥�𝑥𝑥=𝜋𝜋/3

= cos(π/3) = 1/2

11. lim 𝑥𝑥→𝜋𝜋/3

sin 𝑥𝑥 − √32

𝑥𝑥 − 𝜋𝜋3

= �𝑑𝑑𝑑𝑑𝑥𝑥

sin 𝑥𝑥�𝑥𝑥=𝜋𝜋/3

= cos(π/3) = 1/2

12. lim 𝑥𝑥→𝜋𝜋/4

cos 𝑥𝑥 − √22

𝑥𝑥 − 𝜋𝜋4

= �𝑑𝑑𝑑𝑑𝑥𝑥

cos 𝑥𝑥�𝑥𝑥=𝜋𝜋/4

= −sin �π4� = −

√22

13. lim 𝑥𝑥→0

sin 𝑥𝑥𝑥𝑥

= lim 𝑥𝑥→0

sin 𝑥𝑥 − sin 0𝑥𝑥 − 0

= �𝑑𝑑𝑑𝑑𝑥𝑥

sin 𝑥𝑥�𝑥𝑥=0

= cos 0 = 1

14. lim 𝑥𝑥→𝜋𝜋

sin 𝑥𝑥𝑥𝑥 − 𝜋𝜋

= lim 𝑥𝑥→𝜋𝜋

sin 𝑥𝑥 − sin𝜋𝜋𝑥𝑥 − 𝜋𝜋

= �𝑑𝑑𝑑𝑑𝑥𝑥

sin 𝑥𝑥�𝑥𝑥=𝜋𝜋

= cos𝜋𝜋 = −1