# Derivative Of The multiplying Of Functions

## The mathematics lesson, derivatives subject. Finding the derivative of the multiplying of two functions. Product rule in derivates. The subject expression and solved questions.

**Product Rule For Derivatives**

If f(x) and g(x) are two functions and if they have derivatives at point x, the function f.g has a derivative at point x too. The derivative of function f.g is calculated as follows.

[f(x) . g(x)]’ = f’(x) . g(x) + g’(x).f(x)

This equation is called the multiplication rule in derivatives.

**Example:**

f(x) and g(x) are two functions and

f(x) = 5x^{4} + x^{2}

g(x) = 6x^{3}

According to this, calculate the [f(x).g(x)]’.

**Solution:**

Requested from us is we find the derivative of multiplication of functions f(x) and g(x) .

[f(x) . g(x)]’ = f’(x) . g(x) + g’(x).f(x)

Firstly, we will calculate derivative of the function f(x), then we will calculate derivative of the function g(x) and finally we placed these equations in its places in the derivative equality.

The derivative of the function f(x) is

f’(x) = 20x^{3} + 2x

The derivative of the function g(x) is

g’(x) = 18x^{2}

Let we use the multiplication rule

[f(x)*g(x)]’ = [(20x^{3} + 2x) * 6x^{3} ] + [(5x^{4} + x^{2})*18x^{2}]

[f(x)*g(x)]’ = 120x^{6} + 12x^{4} + 90x^{6} + 18x^{4}

[f(x)*g(x)]’ = 210x^{6} + 30x^{4}

We can also do this operation first multiplying the functions each other then taking the derivative of this multiply.

Let we do this question from the second way too

f(x) * g(x) = (5x^{4} + x^{2})(6x^{3})

f(x) * g(x) = 30x^{7} + 6x^{5}

[f(x) * g(x)]’ = 210x^{6} + 30x^{4}

**Example:**

It is given two functions defined in the real numbers cluster.

f(x) = x^{3} – 3x^{2}

g(x) = 3x^{2} – 2x

Calculate the dy/dx operation.

**Solution:**

Requested from us is we find the derivative of multiplying of functions f(x) and g(x) .

First, we will calculate the derivatives of the function f(x) and g(x), then placed these derivatives in placed its places in the equation.

f’(x) = 3x^{2} – 6x

g’(x) = 6x - 2

[f(x) * g(x)]’ = f’(x).g(x) + g’(x) . f(x)

[f(x) * g(x)]’ = (3x^{2} – 6x) . (3x^{2} – 2x) + (6x – 2)(x^{3} – 3x^{2})

[f(x) * g(x)]’ = 9x^{4} – 6x^{3} – 18x^{3} + 12x^{2} + 6x^{4} – 18x^{3} – 2x^{3} + 6x^{2}

[f(x) * g(x)]’ = 15x^{4} – 44x^{3}+ 18x^{2}

A second way

[f(x) * g(x)] = (x^{3} – 3x^{2})( 3x^{2} – 2x)

[f(x) * g(x)] = 3x^{5} – 2x^{4} – 9x^{4} + 6x^{3}

[f(x) * g(x)] = 3x^{5} – 11x^{4 } + 6x^{3}

[f(x) * g(x)]’ =15x^{4} – 44x^{3} + 18x^{2}

**Example:**

A function f(x) has defined as the following

y = | (x^{3} + 2x^{2}) (x + 3) | |

x^{2} + 5x + 6 |

Do the dy/dx operation.

**Solution:**

First, we will separate the numbers in the numerator and denominator to their multipliers .

y = | x^{2}(x + 2)(x + 3) | |

(x + 2)(x + 3) |

y = x^{2}

Now we can calculate its easily.

y’ = 2x

**Example:**

A function h(x) is given as the follows.

h(x) = (5x^{2} + x^{3}) (x^{2} – 2)

How many the value of the derivative of this function for x = 1?

**Solution:**

We can write the function h(x) in f(x).g(x) way.

Suppose that

f(x) = 5x^{2} + x^{3}

g(x) = x^{2} – 2

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

h’(x) = f’(x) . g(x) + g’(x). f(x)

= (10x + 3x^{2})(x^{2} – 2) + 2x(5x^{2} + x^{3})

= 10x^{3} – 20x + 3x^{4} – 6x^{2} + 10x^{3} + 2x^{4}

h’(x) = 5x^{4} + 20x^{3} – 6x^{2} – 20x

h’(1) = 5 + 20 – 6 – 20

= - 1

**Example:**

f(x) = x^{2} – x + 1

g(x) = 2x + 5

A) Calculate the derivate of the [f(x).g(x)]

B) Find the value of this derivative for x = 2.

**Solution:**

A)

[f(x)*g(x)]’ = f’(x) * g(x) + g’(x)*f(x)

f’(x) = 2x – 1

g’(x) = 2

[f(x)*g(x)]’ = (2x – 1)(2x + 5) + 2(x^{2} – x + 1)

= 4x^{2} + 10x – 2x – 5 + 2x^{2} – 2x + 2

[f(x)*g(x)]’ = 6x^{2} + 6x – 3

B) For x = 2

y = 6*(2)^{2} + 6*2 – 3

y = 24 + 12 – 3

y = 33

The second way

f(x) * g(x) = (x^{2} – x + 1)(2x + 5)

f(x) * g(x) = 2x^{3} + 5x^{2} – 2x^{2} – 5x + 2x – 5

f(x) * g(x) = 2x^{3} + 3x^{2} – 3x – 5

[f(x) * g(x)]’ = 6x^{2} + 6x – 3

For x = 2

6.2^{2} + 6.2 – 3

= 24 + 12 – 3

= 33

**The derivative in divide operation**

**The derivatives In Mathematics**

RISE KNOWLEDGE

February 10 2019

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