# The Derivative Of Quotient Of Functions

## The mathematics lesson. Derivatives topic. The quotient rule in derivatives. Derivatives of the functions that in f(x) / g(x) form. The derivative of the division of two functions.

If f(x) and g(x) are two functions and g(x) )≠ 0, the derivative of the f(x) / g(x) is

[f(x)
 ]’ = f’(x) . g(x) – g’(x).f(x) [g(x)]2
g(x)

Example:

A function f(x) has given in the form at following.

f(x) = 1/x

Find the derivative of the function f(x).

Solution:

Let we partition the function f(x) to g(x) and h(x) functions.

Suppose that the function in the numerator is g(x) and the function in the denominator is h(x).

g(x) = 1

g’(x) = 0

h(x) = x

h’(x) =1

 f(x) = g(x) h(x)

 f’(x) = g’(x) . h(x) – h’(x) . g(x) [g(x)]2

 f’(x) = 0 – 1 x2

 f’(x) = – 1 x2

Example:

A function f(x) has given in the form at following.

 f(x) = x – 1 x + 1

Find the derivative of the function f(x).

Solution:

Suppose that the function in the numerator is g(x) , and the function in the denominator is h(x) .

 f(x) = g(x) h(x)

g(x) = x – 1

h(x) = x + 1

g’(x) = 1

h’(x) = 1.

The derivative of division of two function is given in the form below.

 f’(x) = g’(x) . h(x) – h’(x) . g(x) [h(x)]2

 f’(x) = 1.(x + 1) – 1.(x – 1) (x + 1)2

 f’(x) = 2 (x + 1)2

Example:

The functions f(x) and g(x) are defined in the cluster of the real numbers.

f(x) = x2 + 14x + 48

g(x) = x2 + 2x + 48

 Calculate the [ f(x) ]’ g(x)

Solution:

 y = f(x) g(x)

 y = x2 + 14x + 48 x2 + 2x + 48

Let we exam whether f (x) and g (x) functions can be separated into its multipliers.

f(x) = (x + 8)(x + 6)

g(x) = (x – 6)(x + 8)

Both functions can be seperated into its multipliers. Let we simplfy its now.

f(x)
 = (x + 8)(x + 6) (x – 6)(x + 8)
g(x)

 y = (x + 6) (x – 6)

Now, let we calculate its derivates.

 y' = (x + 6)’ . (x – 6) – (x – 6)’ . (x + 6) (x – 6)2

 y' = 1.(x – 6) – 1.(x + 6) (x – 6)2

 y' = x – 6 – x – 6 (x – 6)2

 y’ = - 12 (x – 6)2

Example:

 k(x) = x5 + 2x2 x2 + 3x

Find the value of the k’(2).

Solution:

 k(x) = x^5 + 2x^2 x^2 + 3x

 k(x) = x(x4 + 2x) x(x + 3)

 k(x) = x4 + 2x x + 3

 k’(x) = (x4 + 2x)’ . (x + 3) – (x + 3)’ . (x4 + 2x) (x + 3)2

 k’(x) = (4x3 + 2) . (x + 3) – 1.(x4 + 2x) (x + 3)

 k’(x) = 4x4+ 12x3 + 2x + 6 – x4 – 2x (x + 3)2

 k’(x) = 3x4 + 12x3  + 6 (x + 3)2

 k'(2) = 3.24 + 12.23+ 6 52

k'(2) = 6

Example:

The function f(x) has given in the form at following.

 y = 3x2 + 12x 6x+ 7

 Calculate the dy (1) dx

Solution:

Suppose that

 y = h(x) k(x)

The function hx is

h(x) = 3x2 + 12x

The function k(x) is

k(x) = 6x + 7

Derivative of the function h(x) is

h’(x) = 6x + 12

Derivative of the function k(x) is

k’(x) = 6

The rule of the division in the derivates.

 y' = h’(x) . k(x) – k’(x) . h(x) [k(x)]2

 y' = (6x + 12) . (6x + 7) – 6(3x2 + 12x) (6x + 7)2

 y' = 36x2 + 42x + 72x + 84 – 18x2 – 72x (6x + 7)2

 y' = 18x2 + 42x + 84 (6x + 7)2

dy
 (1) = 18 + 42 + 84 132
dx

dy
 (1) = 144 169
dx

Derivates In Mathematics

RISE KNOWLEDGE

February 5 2019

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