differentiation using product and quoti rule...differentiation using product and quotient rule 1....
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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
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