DIFFERENTIATION USING PRODUCT RULE

Example 1 :

Differentiate x⁵ tan x

Solution :

Let y  =  x⁵ tan x

u  =  x⁵ and 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]

Example 2 :

Differentiate (x²+7x+2) (x³-log x)

Solution :

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)

Example 3 :

Differentiate (x²-1) (x²+2)

Solution :

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

Example 4 :

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

determine the value of (fg)′(2).

Solution :

(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

Example 5 :

Suppose that f and g are functions that are differentiable at x = 1 and that 

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

If h(x)  =  (x²+9)g(x)

Solution :

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

Example 6 :

If

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

determine the value of f′(−7). 

Solution :

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

Example 7 :

Suppose that

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

Solution :

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

Example 8 :

Find all x-coordinates of all points on y = √x (3 - x)² where the tangent is horizontal.

Solution :

y = √x (3 - x)²

Since two functions are multiplied, we have to use product rule to find the derivative.

U =  √x and V = (3 - x)²

U' = 1/2√x and V' = 2(3 - x) (-1) ==> -2(3 - x)

dy/dx = √x(-2(3 - x)) + (3 - x)² (1/2√x)

= -2√x(3 - x) + (3 - x)² (1/2√x)

= [-4x(3 - x) + (3 - x)²] / 2√x

For horizontal tangents, the slope will be equal to 0.

dy/dx = 0

[-4x(3 - x) + (3 - x)²] / 2√x = 0

-4x(3 - x) + (3 - x)² = 0

(3 - x)[-4x + 3 - x] = 0

(3 - x)(-5x + 3) = 0

x = 3 and x = 3/5

So, the required x-coordinates are 3 and 3/5.

Example 9 :

Suppose y = -2x²(x + 4). For what values of x does dy/dx = 10 ?

Solution :

y = -2x²(x + 4)

Let U = -2x² and V = x + 4

U' = -4x and V' = 1

dy/dx =  [-2x² (1) + (x + 4)(-4x)]

= [-2x² - 4x² - 16x]

= -2[3x² + 8x]

When dy/dx = 10

-6x² - 16x = 10

6x² + 16x + 10 = 0

Dividing by 2, we get

3x² + 8x + 5 = 0

3x² + 3x + 5x + 5 = 0

3x(x + 1) + 5(x + 1) = 0

(3x + 5)(x + 1) = 0

x = -1 and x = -5/3

When x = -1 and x = -5/3, the value of dy/dx will be 10.

Example 10 :

Find the gradient of tangent to y = x³√(5 - x²) at x = 1

Solution :

y = x³√(5 - x²)

Let U = x³ and V = √(5 - x²)

U' = 3x² and V' = [1/2√(5 - x²)](-2x)

V' = -x/√(5 - x²)

dy/dx = x³(-x/√(5 - x²)) + √(5 - x²) (3x²)

=  (-x⁴/√(5 - x²)) + 3x²√(5 - x²)

dy/dx at x = 1

= [-1/√(5 - 1²)) + 3(1)²√(5 - 1²)]

= -1/2 + 3(2)

= -1/2 + 6

= 11/2

So, the required slope of the tab slope pf the tangent line is 11/2.

Example 11 :

Find the gradient of tangent to y = x√(1 - 2x) at x = -4

Solution :

y = x√(1 - 2x)

Let U = x and V = √(1 - 2x)

U' = 1 and V' = (1/2√(1 - 2x))(-2)

= -1/√(1 - 2x)

dy/dx = x(-1/√(1 - 2x)) + √(1 - 2x) (1)

= -x/√(1 - 2x) + √(1 - 2x)

dy/dx at x = -4

= 4/√9 + √9

= 4/3 + 3

= 13/3

So, the gradient of the curve at x  = -4 is 13/3.

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