1 solve problems about composite...

1 Solve Problems about Composite Functions1. Differentiate each of the following with respect to x:

(a) f (x) = sin 2x

Solution:

f = sin u u = 2x⇒ f ′ (u) = cos u ⇒ u′ (x) = 2

⇒ f ′ (x) = f ′ (u) × u′ (x) = cos u × 2 = 2 cos 2x

(b) f (x) = cos 2x

Solution:

f = cos u u = 2x⇒ f ′ (u) = − sin u ⇒ u′ (x) = 2

⇒ f ′ (x) = f ′ (u) × u′ (x) = − sin u × 2 = −2 sin 2x

(c) f (x) = tan 2x

Solution:

f = tan u u = 2x

⇒ f ′ (u) =1

cos2 u⇒ u′ (x) = 2

⇒ f ′ (x) = f ′ (u) × u′ (x) =1

cos2 u× 2 =

2cos2 2x

Solutions - Differentiation with Trig of 26

(d) y = sin2 x

Solution:

y = u2 u = sin x

⇒dydu

= 2u ⇒dudx

= cos x

⇒dydx

=dydu×

dudx

= 2u × cos x = 2 sin x cos x

(e) y = cos2 x

Solution:

y = u2 u = cos x

⇒dydu

= 2u ⇒dudx

= − sin x

⇒dydx

=dydu×

dudx

= 2u × − sin x = −2 cos x sin x

(f) y = tan2 x

Solution:

y = u2 u = tan x

⇒dydu

= 2u ⇒dudx

=1

cos2 x

⇒dydx

=dydu×

dudx

= 2u ×1

cos2 x=

2 tan xcos2 x


(g) y = sin3 (2x)

Solution:

y = u3 u = sin v v = 2x

⇒dydu

= 3u2 ⇒dudv

= cos v ⇒dvdx

= 2

⇒dydx

=dydu×

dudv×

dvdx

= 3u2 × cos v × 2

= 6 sin2 (2x) cos (2x)

(h) f (x) = cos2(x2 + 7

)Solution:

f = u2 u = cos v = x2 + 7⇒ f ′ (u) = 2u ⇒ u′ (v) = − sin v ⇒ v′x = 2x

⇒ f ′ (x) = f ′ (u)× ⇒ u′ (v) × v′ (x)= 2u × − sin v × 2x

= −4x cos(x2 + 7

)sin

(x2 + 7

)

(i) y = tan4(2x3 + 4x2 − 6x − 8

)Solution:

y = u4 u = tan v v = 2x3 + 4x2 − 6x − 8

⇒dydu

= 4u3 ⇒dudv

=1

cos2 v⇒

dvdx

= 6x2 + 8x − 6

⇒dydx

=dydu×

dudv×

dvdx

= 4u3 ×1

cos2 v×

(6x2 + 8x − 6

)=

4(6x2 + 8x − 6

)tan3

(2x3 + 4x2 − 6x − 8

)cos2 (

2x3 + 4x2 − 6x − 8)


2 Solve Problems about Products

2. Differentiate each of the following with respect to x:

(a) f (x) = x sin x

Solution:

u = x v = sin x

⇒dudx

= 1 ⇒dvdx

= cos x

⇒ f ′ (x) = x cos x + sin x × 1= x cos x + sin x

(b) f (x) = x2 cos x

Solution:

u = x2 v = cos x

⇒dudx

= 2x ⇒dvdx

= − sin x

⇒ f ′ (x) = x2 × − sin x + cos x × 2x

= −x2 sin x + 2x cos x

(c) f (x) = ex tan x

Solution:

u = ex v = tan x

⇒dudx

= ex ⇒dvdx

=1

cos2 x

⇒ f ′ (x) = ex ×1

cos2 x+ tan x × ex

=ex

cos2 x+ ex tan x


(d) f (x) = ln x sin x

Solution:

u = ln x v = sin x

⇒dudx

=1x

⇒dvdx

= cos x

⇒ f ′ (x) = ln x × cos x + sin x ×1x

= ln x cos x +sin x

x

(e) f (x) =(x2 + 4x + 3

)sin x

Solution:

u = x2 + 4x + 3 v = sin x

⇒dudx

= 2x + 4 ⇒dvdx

= cos x

⇒ f ′ (x) =(x2 + 4x + 3

)× cos x + sin x × (2x + 4)

=(x2 + 4x + 3

)cos x + (2x + 4) sin x

(f) y = 2 cos x sin x

Solution:

u = 2 cos x v = sin x

⇒dudx

= −2 sin x ⇒dvdx

= cos x

⇒dydx

= 2 cos x × cos x + sin x × −2 sin x

= 2 cos2 x − 2 sin x


(g) y = ex ln x sin x

Solution: This is a triple product, but the product rule only deals with two. Thus, weneed to apply the product rule twice. Only the product with sin x is shown here. Trydifferentiating ex ln x yourself and see if you get the answer shown below.

u = ex ln x v = sin x

⇒dudx

=ex

x+ ex ln x ⇒

dvdx

= cos x

⇒dydx

= ex ln x × cos x + sin x ×(ex

x+ ex ln x

)= ex ln x cos x + ex ln x sin x +

ex sin xx

3 Solve Problems about Quotients

3. Differentiate each of the following with respect to x:

(a) f (x) =sin x

x

Solution:

u = sin x v = x

⇒dudx

= cos x ⇒dvdx

= 1

⇒ f ′ (x) =x cos x − 1 sin x

x2

=x cos x − sin x

x2


(b) f (x) =1

tan x

Solution:

u = 1 v = tan x

⇒dudx

= 0 ⇒dvdx

=1

cos2 x

⇒ f ′ (x) =tan x × 0 − 1

cos2 x

tan2 x

=1

tan2 x cos2 x

(c) f (x) =x

cos x

Solution:

u = x v = cos x

⇒dudx

= 1 ⇒dvdx

= − sin x

⇒ f ′ (x) =cos x × 1 − x × − sin x

cos2 x

=cos x + x sin x

cos2 x


(d) f (x) =sin x

1 + cos x

Solution:

u = sin x v = 1 + cos x

⇒dudx

= cos x ⇒dvdx

= − sin x

⇒ f ′ (x) =(1 + cos x) × cos x − sin x × − sin x

(1 + cos x)2

=cos x + cos2 x + sin2 x

(1 + cos x)2

=1 + cos x

(1 + cos x)2

=1

1 + cos x

(e) y =cos2 xsin2 x

Solution:

u = cos2 x v = sin2 x

⇒dudx

= −2 cos x sin x ⇒dvdx

= 2 sin x cos x

⇒dydx

=sin2 x × −2 cos x sin x − cos2 x × 2 sin x cos x

sin4 x

=−2 cos x sin x

(sin2 x + cos2 x

)sin4 x

=−2 cos x sin x

sin4 x

=−2 cos x

sin3 x


(f) y =sin 2x

2 + cos 2x

Solution:

u = sin 2x v = 2 + cos 2x

⇒dudx

= 2 cos 2x ⇒dvdx

= −2 sin 2x

⇒dydx

=(2 + cos 2x) × 2 cos 2x − sin 2x × −2 sin 2x

(2 + cos 2x)2

=4 cos 2x + 2 cos2 2x + 2 sin2 2x

(2 + cos 2x)2

=4 cos 2x + 2(2 + cos 2x)2

(g) y =sin 3x

ex

Solution:

u = sin 3x v = ex

⇒dudx

= 3 cos 3x ⇒dvdx

= ex

⇒dydx

=ex × 3 cos 3x − sin 3x × ex

e2x

=3ex cos 3x − ex sin 3x

e2x

=3 cos 3x − sin 3x

ex


4 Find Higher Derivatives

4. Find the second derivative of each of the following:

(a) f (x) = sin x

Solution:

f ′ (x) = cos xf ′′ (x) = − sin x

(b) f (x) = cos x

Solution:

f ′ (x) = − sin xf ′′ (x) = − cos x

(c) f (x) = tan x

Solution:

f ′ (x) =1

cos2 x= (cos x)−2

f = u−2 u = cos x

f ′ = −2u−3 u′ = − sin x

f ′′ (x) = −2u−3 × − sin x

=2 sin xcos3 x


(d) f (x) = sin (2x)

Solution:

f ′ (x) = 2 cos (2x)f ′′ (x) = −4 sin (2x)

(e) f (x) = cos (3x)

Solution:

f ′ (x) = −3 sin (3x)f ′′ (x) = −9 cos (3x)

(f) f (x) = sin2 x

Solution:

f ′ (x) = 2 sin x cos x

f ′′ (x) = 2 cos2 x − 2 sin x(see product rule)

(g) f (x) = cos3 x

Solution:

f ′ (x) = −3 cos2 x sin x

u = −3 cos2 x v = sin x

⇒dudx

= 6 cos x sin x ⇒dvdx

= cos x

f ′′ (x) = −3 cos2 x × cos x + sin x × 6 cos x sin x

= −3 cos3 x + 6 cos x sin2 x


(h) f (x) = tan2 (2x)

Solution:

f = u2 u = tan v v = 2x

⇒ f ′ = 2u ⇒ u′ =1

cos2 v⇒ v′ = 2

f ′ (x) = 2 tan v ×1

cos2 v× 2

=4 tan (2x)cos2 (2x)

=4 sin (2x)cos3 (2x)

(not necessary, but sin/cos are easier than tan for next step)

u = 4 sin (2x) v = cos3 (2x)

⇒dudx

= 8 cos (2x) ⇒dvdx

= −6 cos2 (2x) sin (2x)

f ′′ (x) =cos3 (2x) × 8 cos (2x) − 4 sin (2x) × −6 cos2 (2x) sin (2x)

cos6 (2x)

=8 cos4 (2x) + 24 sin2 (2x) cos2 (2x)

cos6 (2x)

=8 cos2 (2x) + 24 sin2 (2x)

cos4 (2x)

=1 + 16 sin2 (2x)

cos4 (2x)(not really necessary but looks clean)

5 Solve Problems about Gradients

5. QB T6P1 Q1 Let g (x) = 2x sin x

(a) Find g′ (x).

Solution:

g′ (x) = 2 sin x + 2x cos x


(b) Find the gradient of the graph of g at x = π.

Solution:

g′ (π) = 2 sin π + 2π cos π= 0 − 2π = −2π

6. QB T6P1 Q9 Let h (x).

Find h′ (0)

Solution:

h′ (x) =6 cos x + 6x sin x

cos2 x

⇒ h′ (0) =6 cos 0 + 6 (0) sin 0

cos2 0

=612

= 6

7. QB T6P4 Q20 Consider f (x) = x2 sin x

(a) Find f ′ (x).

Solution:

f ′ (x) = 2x sin x + x2 cos x

(b) Find the gradient of the curve of f at x = π2 .

Solution:

f ′(π

2

)= 2

(π

2

)sin

π

2+

(π

2

)2cos

π

2= π + 0= π


8. QB T6P2 Q4 Let f (x) = e2x cos x, −1 ≤ x ≤ 2.

(a) Show that f ′ (x) = e2x (2 cos x − sin x).

Solution:

u = e2x v = cos x

⇒dudx

= 2e2x ⇒dvdx

= − sin x

⇒ f ′ (x) = e2x × − sin x + cos x × 2e2x

= −e2x sin x + 2e2x cos x

= e2x (2 cos x − sin x)

(b) Find the equation of L.

Solution:

mT = f ′ (0) = e2×0 (2 cos 0 − sin 0)

= e0 (2 − 0)= 2

⇒ mL = −12

f (0) = e2×0 cos 0= 1

y − y1 = m (x − x1)

⇒ y − 1 = −12

(x − 0)

⇒ y = −x2

+ 1


(c) Find the x-coordinate of P.

Solution:

e2x cos x = −x2

+ 1

9. 11M.2.sl.TZ1.8(a) Show that a = 4.

Solution: Amplitude is half the diameter⇒ a = 4.

(b) Show that b = π15 .

Solution:

T = 30

⇒ b =2π30

=π

15


(c) Find these values of t.

Solution:

h′ (t) = −0.5 =4π15

cosπ

15t

⇒ −0.5 ×154π

= cosπ

15t

⇒ −0.5968310... = cosπ

15t

⇒π

15t = 2.210342088 or ⇒

π

15= 4.072843219

⇒ t = 10.55s ⇒ t = 19.45

(d) Determine whether the bucket is underwater at the second value of t.

Solution:

h (19.45) = 4 sin(π

15(19.45)

)+ 2

= −1.211 < 0⇒ under water

6 Solve Problems about Extrema

10. Find the x-coordinate of all extrema of each of the following functions for 0 ≤ x ≤ 2π.

Really need to be on top of radians and unit circle to tackle these questions effectively. Inparticular, note how having a function within a trig function requires a lot of consideration ofextra cycles and reverse cycles of the unit circle to obtain all valid solutions. For sin (bx + c)(or cos), there should be ≈ 2b solutions. Solutions below will skip some aspects of differenti-ation as a time-saver.

(a) cos x

Solution:

f ′ (x) = − sin x = 0⇒ x = 0, π, 2π


(b) sin (2x)

Solution:

f ′ (x) = 2 cos (2x) = 0

⇒ 2x =π

2,

3π2,

5π2,

7π2

⇒ x =π

4,

3π4,

5π4,

7π4

(c) cos (3x)

Solution:

f ′ (x) = −3 sin (3x) = 0⇒ 3x = 0, π, 2π, 3π, 4π, 5π, 6π

⇒ x = 0,π

3,

2π3, π,

4π3,

5π3, 2π

(d) sin(2x + π

2

)Solution:

f ′ (x) = 2 cos(2x +

π

2

)= 0

⇒ 2x +π

2= −

π

2,

pi2,

3π2,

5π2,

7π2,

9π2

⇒ 2x = −π, 0, π, 2π, 3π, 4π

⇒ x =��−π

2, 0,

π

2, π,

3π2, 2π

(e) cos(

12 x − π

)Solution:

f ′ (x) = −12

sin(12

x − pi)

= 0

⇒12

x − π = −π, 0, π, 2π, 3π

⇒12

x = ��−2π, 0, 2π,��4π,��6π


(f) 2 sin (4x + 2π)

Solution:

f ′ (x) = 8 cos (4x + 2π) = 0

⇒ 4x + 2π = −π

2,π

2,

3π2,

5π2,

7π2,

9π2,

11π2,

13π2,

15π2,

17π2,

19π2,

21π2

⇒ 4x = −5π2,−

3π2,−π

2,π

2,

3π2,

5π2,

7π2,

9π2,

11π2,

13π2,

15π2,

17π2

⇒ x =��

−5π8,��

−3π8,��−π

8,π

8,

3π8,

5π8,

7π8,

9π8,

11π8,

13π8,

15π8,��17π

8

11. QB T6P3 Q21(a) Use the graph to write down the value of a, c, and d.

Solution:

a =12

(12 − −4) = 8c = 4 −14

wavelength = 2d = 12 − a = 4

(b) Show that b = π4 .

Solution:

b =2πT

=2π8

=π

4

(c) Find f ′ (x).

Solution:

f (x) = 8 sin(π

4(x − 2)

)+ 4 = 8 sin

(π

4x −

π

2

)+ 4⇒ f ′ (x) = 2π cos

(π

4(x − 2)

)


(d) Find the x-coordinate of R.

Solution:

2π cos(π

4(x − 2)

)= −2π

⇒ cos(π

4(x − 2)

)= −1

⇒π

4(x − 2) = −π, π, 3π...

⇒π

4x −

π

2= −π, π, 3π...

⇒π

4x = −

π

2,

3π2,

7π2...

⇒ x = ��−2, 6,��14

7 Solve Problems about Points of Inflexion

12. Find the x-coordinate of all points of inflexion of each of the following functions for 0 ≤ x ≤2π.

(a) cos x

Solution:

f ′′ (x) = − cos x

⇒ x =π

2,

3π2

(b) tan x

Solution:

f ′′ (x) =2 sin xcos3x

⇒ x = 0, π, 2π

(c) sin (2x)

Solution:

f ′′ (x) = −4 sin (2x)

⇒ x = 0,π

2, π,

3π2, 2π


(d) cos (3x)

Solution:

f ′′ (x) = −9 cos (3x)

⇒ x =π

6,π

2,

5π6,

7π6,

3π2,

11π6

(e) tan (2x)

Solution:

f ′′ (x) =8 sin (2x)cos3 (2x)

⇒ x = 0,π

2, π,

3π2, 2π

(f) sin(2x + π

2

)Solution:

f ′′ (x) = −4 sin(2x +

π

2

)⇒ x =

π

4,

3π4,

5π4,

7π4

(g) cos(

12 x − π

)Solution:

f ′′ (x) = −14

cos(12

x − π)

⇒ x = π

(h) 2 sin (4x + 2π)

Solution:

f ′′ (x) = −32 sin (4x + 2π)

⇒ x = 0,π

4,π

2,

3π4, π,

5π4,

3π2,

7π4, 2π


13. 10M.1.sl.TZ1.9(a) Use the quotient rule to show that f ′ (x) = − 1

sin2 x.

Solution:

u = cos x v = sin x

⇒dudx

= − sin x ⇒dvdx

= cos x

⇒ f ′ (x) =(sin x) (−sinx) − (cos x) (cos x)

sin2 x

=− sin2 x − cos2 x

sin2 x

=−1

(sin2 x + cos2 x

)sin2 x

= −1

sin2 x

(b) Find f ′′ (x).

Solution:

f ′ (x) = sin−2 x

⇒ f ′′ (x) =2 cos xsin3 x

(c) Find the values of p and q.

Solution:

f ′(π

2

)= −

1sin2 π

2

= −1

f ′′(π

2

)=

2 cos π2

sin3 π2

= 0

(d) Use the information in the table to explain why there is a point of inflexion on the graphof f where x = π

2 .

Solution: f ′′ (x) has a root at π2 and f changes from concave up to concave down at

π2 , indicating that it is a point of inflexion.


8 Solve Optimization Problems

14. 11M.2.sl.TZ2.10(a) Show that the area of the window is given by y = 4 sin θ + 2 sin 2θ.

Solution:

Area =a + b

2h a = 2

y =2 + 2 + 4 cos θ

2× 2 sin θ h = 2 sin θ

=4 + 4 cos θ

2× 2 sin θ b = 2 + 4 cos θ

= (2 + 2 cos θ) (2 sin θ)= 4 sin θ + 2 (2 cos θ sin θ)= 4 sin θ + 2 sin 2θ

(b) Zoe wants a window to have an area of 5 m2. Find the two possible values of θ.

Solution:

4 sin θ + 2 sin 2θ = 5

From graph, θ ≈ 0.85, 1.25


(c) Find all possible values for A.

Solution: From graph, it looks like approximately 4 < A < 5.2.

y(π

2

)= 4 sin

(π

2

)+ 2 sin

(2π

2

)= 4 (1) + 2 (0) = 4

, so minimum value is 4.For maximum value, need derivative = 0:

y′ (θ) = 4 cos θ + 4 cos 2θ = 0⇒ cos θ + cos 2θ = 0

⇒ cos θ + 2 cos2 θ − 1 = 0 cos 2θ = 2 cos2 θ − 1

⇒ 2 cos2 θ + cos θ − 1 = 0

⇒ 2u2 + u − 1 = 0 u = cos θ⇒ (2u − 1) (u + 1) = 0

⇒ u =12

⇒ u = −1

⇒ cos θ =12

⇒ cos θ = −1

⇒ θ =π

3⇒ θ =�π

y(π

3

)= 4 sin

(π

3

)+ 2 sin

(2π

3

)≈ 5.20

, so possible values for A are 4 ≤ A < 5.20

15. 08M.1.sl.TZ2.10(a) Find the area of the triangle OPB in terms of θ.

Solution:

A =12

ab sin θ

=12

(2) (2) sin θ

= 2 sin θ


(b) Explain why the area of the triangle OPA is the same as the area of triangle OPB.

Solution: They have the same base length (2) and the same height (as their uppervertex is in the same place).

(c) Show that S = 2 (π − 2 sin θ).

Solution:

S =12πr2 − 2 (2 sin θ)

=12π22 − 4 sin θ

= 2π − 4 sin θ= 2 (π − 2 sin θ)

(d) Find the value of θ when S is a local minimum, justifying that it is a local minimum.

Solution:

S ′ (θ) = −4 cos θ = 0⇒ cos θ = 0

⇒ θ =π

2

S(π

2

)= 2

(π − 2 sin

(π

2

))= 2 (π − 2)≈ 2.28

(e) Find a value of θ for which S has its greatest value.

Solution:

θ = 0, π


9 Solve Kinematic Problems

16. 12N.2.sl.TZ0.7(a) Sketch the graph of s.

Solution:

(b) Find the maximum velocity of the particle.

Solution:

⇒ v (t) = −2t cos t + 2 cos t⇒ v′ (t) = −2t cos t − 2 sin t − 2 sin t = 0

⇒ −2t cos t − 4 sin t = 0

⇒ t − 2sin tcos t

= 0

⇒ t − 2 tan t = 0

This is not solvable with our level of mathematics, so we try a graph instead:


Thus vmax ≈ 10.2ms−1

17. 12M.1.sl.TZ1.10(a) Find s′ (t)

Solution:

s′ (t) = 1 − 2 cos (2t)

(b) Find the other value.

Solution:

1 − 2 cos (2t) = 0

⇒ cos (2t) =12

⇒ 2t =π

3,

5π3

⇒ t =��π

6,

5π6

(c) Show that s′ (t) > 0 between these two values of t.

Solution:

1 − 2 cos (2π)= 1 − 2 (−2)= 5 > 0


1 solve problems about composite...

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