PRACTICE PROBLEMS ON DIFFERENTIATION USING PRODUCT RULE

The product rule is the formal rule for differentiating problems where one function is multiplied by another.

(UV)' = UV' + VU'

(UVW)' = UVW'+UV'W+U'VW

Differentiate the following  :

(1)  x5 tan x

(2)  (x2+7x+2) (x3-log x)

(3)    (x2-1) (x2+2)

(4)  If f(2)  =  −8, f′(2)  =  3, g(2)  =  17 and g′(2)  =  −4 

determine the value of (fg)′(2).

(5)  f(1)  =  2, f′(1)  =  −1, g(1)  =  −2, and g′(1)  =  3. Find h′(1). If h(x)  =  (x2+9)g(x)

(6)  f(x)  =  x3g(x), g(−7)  =  2, g′(−7)  =  −9  determine the value of f′(−7). 

(7)  Differentiate y  =  xexsinx

(8)  Differentiate  y  =  xexsinx

(9)  If f(x) = (x2-10)(x2-6x+2) what is the value of f′(6)

(10)  Differentiate sin2x cos 3x

(11)  Suppose f(x) = x2 + (b + 1) x+ 2c, f(2) = 4, and f'(1) = 2. Find the constants b and c.

(12)  The cost of producing x toasters each week is given by C = 1785 + 3x + 0:002x2 dollars. Find dC / dx and interpret its meaning.

Detailed Answer Key

1.  Answers :

Let y  =  x5 tan x

u  =  xand u'  =  5x4 

v = tan x and v'  =  sec2 x

Using the formula 

(UV)' = UV' + VU'

=  (x5)sec2x+(tan x)(5x4)

x5sec2x+5x4tan x

=  x4[xsec2x+tanx]

2.  Answers :

Let y  =  (x2+7x+2) (x3-log x)

u  =  x2+7x+2 and  u' = 2x+7

v  =  x3 - log x and v' = 3x2 - (1/x) 

Formula of product rule

           = (x2+7x+2)[3x2 - (1/x)] + (x3 - log x)(2x+7)

3.  Answer :

Let y  =  (x2-1) (x2+2)

u  =  x2-1 and u'  =  2x

v = x2+2 and v'  =  2x

Using the formula, we get

(UV)' = UV' + VU'

=  (x2-1)(2x)+(x2+2)(2x)

=  2x3-2x+2x3+4x

=  4x3+2x

Another Method :

(x2-1) (x2+2)

=  x2(x2) + 2x2 - 1(x2) - 1(2)

=  x4+2x2-x2-2

=  x4+x2-2

Differentiating

=  4x3 + 2x - 0

=  4x³ + 2x

4.  Answer :

If f(2)  =  −8, f′(2)  =  3, g(2)  =  17 and g′(2)  =  −4 

determine the value of (fg)′(2).

(fg)'(x)  =  f(x) g'(x)+f'(x)g(x)

(fg)'(2)  =  f(2) g'(2)+f'(2)g(2)

  =  -8(-4) + 17(3)

  =  32 + 51

  =  83

5.  Answer :

f(1)  =  2, f′(1)  =  −1, g(1)  =  −2, and g′(1)  =  3. Find h′(1).

If h(x)  =  (x2+9)g(x)

Given that :

h(x)  =  (x2+9)g(x)

Let f(x)  =  x2+9

h'(x)  = (x2+9)g'(x) + g(x) d(x2+9)

h'(x)  =  (x2+9)g'(x) + g(x) (2x)

h'(1)  =  (12+9)g'(1) + g(1) 2(1)

  =  10(3)+(-2)2

  =  30 - 4

  =  26

6.  Answer :

f(x)  =  x3g(x), g(−7)  =  2, g′(−7)  =  −9 

determine the value of f′(−7). 

Given that :

f(x)  =  x3g(x)

f'(x)  =  x3g'(x) + g(x) d(x3)

f'(x)  =  x3g'(x) + 3x2g(x)

f'(-7)  =  (-7)3g'(-7) + 3(-7)2g(-7)

f'(-7)  =  -343(2) + 3(49)(-9)

f'(-7)   =  -686+1323

f'(-7)  =  637

7. Answer :

f(π/4)  =  −4 and  f′(π/4)  =  2, and let  g(x)  =  f(x) sinx.'

g(x)  =  f(x) sinx

g'(x)  =  f(x) d(sinx) + sinx f'(x)

g'(x)  =  f(x) cosx + sinx f'(x)

g'(π/4)  =  f(π/4) cosπ/4 + sinπ/4 f'(π/4)

=  -4(1/√2) + (1/√2)2

=  -2/√2

=  -√2

8. Answer :

Let y  =  xexsinx

(uvw)' =  uvw'+uv'w+u'vw  ---(1)

u  =  x and u'  =  1

v  =  ex and v'  =  ex

w  =  sin x and w'  =  cos x

By applying the values in (1), we get

=  xex cosx + xex sinx + 1ex (sinx)

=  e(x cosx + x sinx + sinx)

9.  Answer :

If f(x) = (x2-10)(x2-6x+2) what is the value of f′(6)

f'(x) = (x2-10)d(x2-6x+2) + (x2-6x+2) d(x2-10)

f'(x)  =  (x2-10)(3x-6) + (x2-6x+2) (2x)

f'(6)  =  (62-10)(3(6)-6) + (62-6(6)+2) (2(6))

f'(6)  =  (36-10)(12) + (36-36+2) 12

f'(6)  =  (26)(12) + 2

f'(6)  =  312+12

f'(6)  =  324

10.   Answer :

Let y  =  sin2x cos 3x

u  =  sin 2x, u'  =  2 cos 2x

v  =  cos 3x, v'  =  -3 sin 3x

Using product rule, we get

sin 2x(-3 sin 3x) + cos 3x(sin 2x)

sinx2x(-3 sin 3x) +  cos 3x(sin 2x)

So, the differentiation of the given function is 

-3(-3 sin 3x) +  cos 3x(sin 2x)

11.   Answer :

Suppose f(x) = x2 + (b + 1) x + 2c, f(2) = 4, and f'(1) = 2. Find the constants b and c.

f(x) = x2 + (b + 1) x + 2c

f(2) = 4

Applying x = 2, we get

f(2) = 22 + (b + 1) (2) + 2c

4 + 2b + 2 + 2c = 4

2b + 2c = -2

b + c = -1-----(1)

f'(x) = 2x + (b + 1) (1) + 0

f'(x) = 2x + b + 1

f'(1) = 2

2 = 2(1) + b + 1

2 = 2 + b + 1

b = -1

Applying the value of b in (1), we get

-1 + c = -1

c = -1 + 1

c = 0

So, the values of b and c are -1 and 0 respectively.

12.   Answer :

C = 1785 + 3x + 0.002x2 dollars

dC/dx = 0 + 3(1) + 0.002(2x)

= 3 + 0.004x

Cost change = 3 + 0.004x

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