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)





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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