οΏ½ οΏ½ οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½ οΏ½ οΏ½ οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½
3οΏ½ οΏ½ οΏ½οΏ½οΏ½ οΏ½ οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½ οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½
5. οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½ οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½
οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½
οΏ½ οΏ½ οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½ οΏ½ οΏ½ οΏ½οΏ½ οΏ½ οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½
οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½
οΏ½ οΏ½οΏ½οΏ½
οΏ½
οΏ½οΏ½οΏ½οΏ½
οΏ½ οΏ½οΏ½οΏ½οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½
οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½
οΏ½ οΏ½οΏ½οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½οΏ½οΏ½
οΏ½
οΏ½οΏ½οΏ½
οΏ½οΏ½ οΏ½οΏ½οΏ½ οΏ½οΏ½16.οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½ οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½
οΏ½ οΏ½οΏ½οΏ½
οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½
οΏ½ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½20.οΏ½ οΏ½οΏ½ οΏ½ οΏ½οΏ½
οΏ½οΏ½οΏ½οΏ½οΏ½
οΏ½
1. 2.
4.
6.
7.8.
9.10.
11. 12.
13. 14.
15.
17. 18.
19.
ππππππππ
= 2(ππ + 1)(2ππ2 + ππ + 1) ππππππππ
=ππ + 12βππ
+ βππ + 2
ππππππππ
= 3ππ2 ππππππππ
= (ππ7 + 15)2(22ππ7 + 15)
ππππππππ
= ππ(ππ + 7)2(5ππ + 14) ππππππππ
=32ππ8 + 49ππ6
β4ππ2 + 7
ππππππππ
=3ππ2 + 4
2βππ
ππππππππ
=2ππ3
βππ4 β 1
ππππππππ
= 2(2ππ β 1)3(10ππ + 3) ππππππππ
= 1
ππππππππ
=1
(ππ + 1)2ππππππππ
=3ππ2 β 2ππ(3ππ β 1)2
ππππππππ
=20ππ4 + 59ππ2 β 14
(5ππ2 + 2)2ππππππππ
=4ππ(ππ2 β 1)2(ππ2 + 2)
(ππ2 + 1)2
ππππππππ
=9ππ8βππ2 β 1 β ππ(ππ9
2β 1)
βππ β 1ππ2 β 1
ππππππππ
=β4ππ4 + 24ππ
(ππ3 + 3)2
ππππππππ
=52
52ππ + 3ππ2
(βππ + 1)2ππππππππ
=β2
(ππ β 1)2
ππππππππ
=2ππ
(ππ2 + 4)2ππππππππ
=
12β βππ + 7
οΏ½ππ + 72οΏ½
24βππ
Differentiation Using Product and Quotient Rule 1
.
21. Find the horizontal tangents of: π¦ = π₯4 β 2π₯2 + 2
Horizontal tangents occur when slope = zero.
π¦β² = 4π₯3 β 4π₯ = 0
π₯(π₯ + 1)(π₯ β 1) = 0
π₯ = 0, β1, 1
πππππ‘π : (0,2) (β1,1) (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 = βπ₯
1+π₯2 at the point (1, Β½).
slope of the tangent line at (1, Β½) :
tangent line at (1, Β½):
π¦ β 1 = β1
4(π₯ β 1) β π¦ = β
1
4π₯ +
3
4
normal line at (1, Β½):
π¦ β1
2= 4(π₯ β 1) β π¦ = 4π₯ β
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:
ππ¦
ππ₯= 12(π₯β1)β² = β
12
π₯2
Let the x-coordinate of one of the points in question be π.
Slope of the tangent line at that point is β12
π2 , and that has to be equal to the slope of line
3x + y = 0
β12
π2 = β3 ππ π = Β±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
__ __ _____
π
ππ₯[π(π₯)π(π₯)] = πβ²(π₯)π(π₯) + π(π₯)πβ²(π₯)
π
ππ₯[π(π₯)
π(π₯)] =
πβ²(π₯)π(π₯) β π(π₯)πβ²(π₯)
π2(π₯)
πβ²(π₯) = 6(π₯3 β 3) + (6π₯ + 5)(3π₯2) ββ²(π₯) = 2 sin π‘ + 2 π‘ cos π‘ + 2π‘ cos π‘ β π‘2 sin π‘
πβ²(π₯) = 4π₯ cot π₯ β 2π₯2 ππ π2π₯
πβ²(π₯) =(sin π₯ + 1)(1 + π ππ2π₯) β (π₯ + tan π₯)(cos π₯)
=
(sin π₯ + 1)2
sin π₯ β π ππ2π₯ + 1 + π ππ2π₯ β π₯ πππ π₯ (sin π₯ + 1)2
(2π₯ β 1)(π₯2 + 1) β (π₯2 β π₯ β 3)(2π₯)πβ²(π₯) = [
(π₯2 + 1)2
π₯2 β π₯ β 3] [π₯2 + π₯ + 1] + [
π₯2 + 1] [2π₯ + 1]
πβ²(π₯) = π ππ2π₯ β sin π₯ + tan π₯ β cos π₯
= sec π₯ β tan π₯ + sin π₯
= 24π₯3 + 15π₯2 β 18 = 2 sin π‘ + 4 π‘ cos π‘ β π‘2 sin π‘
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 .
πβ²(π₯) =(β sin π₯)(π₯2) β (cos π₯)(2π₯)
(π₯2)2
=βπ₯2 sin π₯ β 2π₯ cos π₯
π₯4
= (0 β 1 β cos π₯
π ππ2π₯) (
1
sin π₯) + (
1
sin π₯) (
0 β 1 β cos π₯
π ππ2π₯)
= β2cot π₯
π ππ2π₯
ββ²(π₯) =π
ππ₯[(
1
sin π₯) (
1
sin π₯)]
π (π
, 1) 4
πβ²(π₯) = cos π₯(sin π₯ + cos π₯) + sin π₯(cos π₯ β sin π₯)
= 2 sin π₯ cos π₯ + πππ 2π₯ β π ππ2π₯
= 2 (β2
2) (
β2
2) + (
β2
2)
2
β (β2
2)
2
= 1
π‘ππππππ‘ ππππ ππ‘ π (π
4, 1) : π¦ = π₯ + (1 β
π
4)
π₯ β πππ‘πππ πππ‘πππ: (π₯ β 1)(π₯2 + 1) = 0 β΄ π(1,0)
πβ²(π₯) = (1)(π₯2 + 1) + (π₯ β 1)(2π₯) β΄ πβ²(1) = 2
π‘ππππππ‘ ππππ ππ‘ π(1,0): π¦ = 2π₯ β 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
πππ ππππππππ πππππ βππ£π π‘βπ π πππ π ππππ
π ππππ ππ π‘βπ π‘πππππ‘ ππππ ππ β1
2
π¦β² =(1)(π₯ β 1) β (π₯ + 1)(1)
(π₯ β 1)2
=β2
(π₯ β 1)2
β΄ β2
(π₯ β 1)2= β
1
2 β (π₯ β 1)2 = 4 β (π₯ β 1) = Β± 2 β π₯ = β1 & π₯ = 3
π‘ππππππ‘ ππππ ππ‘ (β1,0) ππ (3,2): π¦ = β1
2π₯ β
1
2 & π¦ = β
1
2π₯ +
7
2
πβ²(π₯) =(3)(π₯ + 2) β (3π₯)(1)
(π₯ + 2)2
=6
(π₯ + 2)2
πβ²(π₯) =(5)(π₯ + 2) β (5π₯ + 4)(1)
(π₯ + 2)2
=6
(π₯ + 2)2
π(π₯) =π₯ + 2
=π₯ + 2
=3π₯ 3(π₯ + 2) β 6 β6
π₯ + 2+ 3
π(π₯) =5π₯ + 4
π₯ + 2=
5(π₯ + 2) β 6
π₯ + 2=
β6
π₯ + 2+ 5
πΌπ πβ²(π₯) = πβ²(π₯), π‘βππ π(π₯)πππ π(π₯)ππππ¦ ππππππ ππ¦ π ππππ π‘πππ‘. πππππ:
____
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
πβ²(π₯) =π
ππ₯[
1
cos π₯] =
(0)(cos π₯) β (1)(β sin π₯)=
sin π₯
πππ 2π₯
πβ²(π₯) =π
ππ₯[
1
sin π₯] =
πππ 2π₯
(0)(sin π₯) β (1)(cos π₯)
π ππ2π₯= β
cos π₯
π ππ2π₯
π ππ π₯= β
cos π₯
πππ 2π₯ π ππ2π₯
π ππ3π₯ = βπππ 3π₯ β sin π₯ = β cos π₯
π₯ =3π
4 & π₯ =
7π
4
πππ π€πππ π€πππ π£πππ¦
πβ²(2) = 0 πβ²(2) = β4 πβ²(2) = β10 πβ²(2) = 28
____
πππππ‘ ππ πππ‘ππππππ‘: π + π + π = 0
ππππβ πππ π ππ π‘βπππ’πβ π‘βπ πππππ‘ (2,7): 4π + 2π + π = 7
π‘ππππππ‘ ππππ ππ‘ (2,7) βππ π‘βπ π ππππ ππ 10: πβ²(π₯) = 2ππ₯ + π β 4π + π = 10
π + π + π = 04π + 2π + π = 74π + π = 10
π = 3 π = β2 π = β1
π(π₯) = 3π₯2 β 2π₯ β 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 ?
πβ²(1) ππ π‘βπ π ππππ ππ π‘ππππππ‘ ππππ ππ‘ π₯ = 1. πΌπ‘ ππ πππππ‘ππ£π ππππππππππ ππ π‘βπ ππππππ ππππ π ππππ:
ππ‘ = β1
ππ
ππ =2 β 1
=1
21 β (β1)
πβ²(1) = ππ‘ = β2
13. If y x sin x , find3
3d ydx
π3π¦
ππ₯3= β3 sin π₯ β π₯ cos π₯
π¦ = cos π₯
π¦β² = βπ ππ π₯
π¦β²β² = β cos π₯
π¦β²β²β² = π ππ π₯
π¦(4) = cos π₯ = π¦
999 = 249 β 4 + 3
π¦(999) = π¦β²β²β² = π ππ π₯
π¦ = π ππ π₯
π¦β² = cos π₯
π¦β²β² = β π ππ π₯
π¦β²β²β² = βπππ π₯
π¦(4) = π ππ π₯ = π¦
725 = 181 β 4 + 1
14. Find the following
(a) x999
999d cosdx
(c) x725
725d sindx
π¦(999) = π¦β² = πππ π₯
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
ππ₯[π ππ 2π₯] =
π π[2 sin π₯ cos π₯]
ππ₯
= 2(cos π₯)(cos π₯) + 2(sin π₯)(β sin π₯)
= 2(πππ 2π₯ β π ππ2π₯)
= 2 cos 2π₯
π
ππ₯[cos 2π₯] =
π
ππ₯[πππ 2π₯ β π ππ2π₯]
=π
[(πππ π₯)(πππ π₯) β (π ππ π₯)(π ππ π₯)] ππ₯
= [(βπ ππ π₯)(πππ π₯) + (πππ π₯)(β π ππ π₯)] β [(πππ π₯)(π ππ π₯) + (π ππ π₯)(πππ π₯)]
= β2 sin π₯ cos π₯ β 2 sin π₯ cos π₯ = β sin 2π₯ β sin 2π₯
= β2 sin 2π₯
π. π
ππ₯[cos 2π₯]
2
SOLUTIONS
1. lim ββ0
(π₯π₯ + β)2 β π₯π₯2
β=
πππππ₯π₯
π₯π₯2 = 2π₯π₯
2. lim ββ0
tan (π₯π₯ + β) β tanββ
=πππππ₯π₯
tan π₯π₯ =1
π π π π π π 2 π₯π₯
3. lim ββ0
(π₯π₯ + β)4 β π₯π₯4
β=
πππππ₯π₯
π₯π₯4 = 4π₯π₯3
4. lim ββ0
2(π₯π₯ + β)2 + 3(π₯π₯ + β) β 2π₯π₯2 β 3π₯π₯β
=πππππ₯π₯
(2π₯π₯2 + 3π₯π₯) = 4π₯π₯ + 3
5. lim ββ0
sin[2(π₯π₯ + β)]β sin 2π₯π₯β
=πππππ₯π₯
sin 2π₯π₯ = 2 cos 2π₯π₯
6. lim ββ0
cos(π₯π₯ + β)2 β cos π₯π₯2
β=
πππππ₯π₯
cos π₯π₯2 = β2π₯π₯ sin π₯π₯2
7. lim π₯π₯β3π₯π₯2β9π₯π₯β3
= lim π₯π₯β3π₯π₯2β32
π₯π₯β3= οΏ½ ππ
πππ₯π₯ π₯π₯2οΏ½
π₯π₯=3= 6
8. lim π₯π₯β5
π₯π₯2 β 52
π₯π₯ β 5= οΏ½
πππππ₯π₯
π₯π₯2οΏ½π₯π₯=5
= 10
9. lim π₯π₯β5
π₯π₯2 β 25π₯π₯ β 5
= lim π₯π₯β5
π₯π₯2 β 52
π₯π₯ β 5= οΏ½
πππππ₯π₯
π₯π₯2οΏ½π₯π₯=5
= 10
10. lim π₯π₯βππ/3
sin π₯π₯ β sinππ3π₯π₯ β ππ
3= οΏ½
πππππ₯π₯
sin π₯π₯οΏ½π₯π₯=ππ/3
= cos(Ο/3) = 1/2
11. lim π₯π₯βππ/3
sin π₯π₯ β β32
π₯π₯ β ππ3
= οΏ½πππππ₯π₯
sin π₯π₯οΏ½π₯π₯=ππ/3
= cos(Ο/3) = 1/2
12. lim π₯π₯βππ/4
cos π₯π₯ β β22
π₯π₯ β ππ4
= οΏ½πππππ₯π₯
cos π₯π₯οΏ½π₯π₯=ππ/4
= βsin οΏ½Ο4οΏ½ = β
β22
13. lim π₯π₯β0
sin π₯π₯π₯π₯
= lim π₯π₯β0
sin π₯π₯ β sin 0π₯π₯ β 0
= οΏ½πππππ₯π₯
sin π₯π₯οΏ½π₯π₯=0
= cos 0 = 1
14. lim π₯π₯βππ
sin π₯π₯π₯π₯ β ππ
= lim π₯π₯βππ
sin π₯π₯ β sinπππ₯π₯ β ππ
= οΏ½πππππ₯π₯
sin π₯π₯οΏ½π₯π₯=ππ
= cosππ = β1