derivative of (6x+8x^2)/((3+8x)^2)

1) Let's perform the differentiation:

\frac{d}{d x} (\frac{6 x + 8 x^{2}}{{(3 + 8 x)}^{2}})

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{6 x + 8 x^{2}}{{(3 + 8 x)}^{2}}

\frac{2 x \cdot (3 + 4 x)}{{(3 + 8 x)}^{2}}

\frac{d}{d x} (\frac{2 x \cdot (3 + 4 x)}{{(3 + 8 x)}^{2}})

We differentiate

p1

2 x \cdot (3 + 4 x)

and

q2

{(3 + 8 x)}^{2}

using:

\frac{d}{d x} (\frac{p1}{q2}) = \frac{\frac{d}{d x} (p1) \cdot q2 - p1 \cdot \frac{d}{d x} (q2)}{{q2}^{2}}

2) Let's differentiate

p1

\frac{d}{d x} (2 x \cdot (3 + 4 x))

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (2 x \cdot (3 + 4 x))

2 \cdot \frac{d}{d x} (x \cdot (3 + 4 x))

2 \cdot \frac{d}{d x} (x \cdot (3 + 4 x))

⇥We differentiate

u3

x

and

v4

3 + 4 x

using:

\frac{d}{d x} (u3 \cdot v4) = \frac{d}{d x} (u3) \cdot v4 + u3 \cdot \frac{d}{d x} (v4)

⇥2.1) Let's differentiate

u3

\frac{d}{d x} (x)

⇥ⓘ The derivative of x is 1:

\frac{d}{d x} (x)

1

⇥

1

⇥2.2) Let's differentiate

v4

\frac{d}{d x} (3 + 4 x)

⇥ⓘ The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (3 + 4 x)

0 + \frac{d}{d x} (4 x)

⇥

0 + \frac{d}{d x} (4 x)

⇥ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (4 x)

\frac{d}{d x} (4 x)

■A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (4 x)

4

⇥

4

⇥2.3) Let's evaluate the resulting derivative (

u3 \cdot v4

)':

1 \cdot (3 + 4 x) + x \cdot 4

⇥

3 + 4 x + x \cdot 4

⇥ⓘ Like terms are grouped by adding their coefficients:

4 x + x \cdot 4

8 x

⇥

3 + 8 x

⇥2.4) Then you can resume the differentiation:

ⅰ

\frac{d}{d x} (x \cdot (3 + 4 x))

3 + 8 x

2 \cdot (3 + 8 x)

ⓘ Expand by multiplying each term inside the parentheses:

2 \cdot (3 + 8 x)

6 + 16 x

6 + 16 x

3) Let's perform the differentiation

q2

\frac{d}{d x} ({(3 + 8 x)}^{2})

Apply the differentiation rule:

\frac{d}{d x} (f^{n}) = n \cdot \frac{d}{d x} (f) \cdot f^{n - 1}

with f =

3 + 8 x

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} ({(3 + 8 x)}^{2})

2 \cdot \frac{d}{d x} (3 + 8 x) \cdot (3 + 8 x)

2 \cdot \frac{d}{d x} (3 + 8 x) \cdot (3 + 8 x)

ⓘ Expand by multiplying each term inside the parentheses:

2 \cdot \frac{d}{d x} (3 + 8 x) \cdot (3 + 8 x)

6 \cdot \frac{d}{d x} (3 + 8 x) + 16 \cdot \frac{d}{d x} (3 + 8 x) \cdot x

■The derivative of a sum is the sum of the derivatives:

\frac{d}{d x} (3 + 8 x)

0 + \frac{d}{d x} (8 x)

6 \cdot (0 + \frac{d}{d x} (8 x)) + 16 \cdot (0 + \frac{d}{d x} (8 x)) \cdot x

ⓘ Adding 0 changes nothing:

0 + \frac{d}{d x} (8 x)

\frac{d}{d x} (8 x)

■A constant factor stays in front of the derivative, (k*u)' = k*u':

\frac{d}{d x} (8 x)

8

6 \cdot 8 + 16 \cdot 8 x

ⓘ Multiplications are done before additions and subtractions:

6 \cdot 8

48

16 \cdot 8

(10 + 6) \cdot 8

10 \cdot 8 + 6 \cdot 8

80 + 48

128

48 + 128 x

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{(6 + 16 x) \cdot {(3 + 8 x)}^{2} - 2 x \cdot (3 + 4 x) \cdot (48 + 128 x)}{{({(3 + 8 x)}^{2})}^{2}}

ⓘ What the terms have in common is factored out, so that the fraction can be simplified:

\frac{(6 + 16 x) \cdot {(3 + 8 x)}^{2} - 2 x \cdot (3 + 4 x) \cdot (48 + 128 x)}{{({(3 + 8 x)}^{2})}^{2}}

\frac{(8 x + 3) \cdot ((8 x + 3) \cdot (16 x + 6) - 32 x \cdot (3 + 4 x))}{{({(3 + 8 x)}^{2})}^{2}}

■A power of a power multiplies the exponents, (a^n)^m = a^(n*m):

{({(3 + 8 x)}^{2})}^{2}

{(3 + 8 x)}^{2 \cdot 2}

\frac{(8 x + 3) \cdot ((8 x + 3) \cdot (16 x + 6) - 32 x \cdot (3 + 4 x))}{{(3 + 8 x)}^{2 \cdot 2}}

ⅰ

2 \cdot 2

4

\frac{(8 x + 3) \cdot ((8 x + 3) \cdot (16 x + 6) - 32 x \cdot (3 + 4 x))}{{(3 + 8 x)}^{4}}

We simplify the fraction:

\frac{8 x + 3}{{(3 + 8 x)}^{4}}

by the common factor:

8 x + 3

to get the new fraction:

\frac{1}{{(3 + 8 x)}^{3}}

\frac{(8 x + 3) \cdot (16 x + 6) - 32 x \cdot (3 + 4 x)}{{(3 + 8 x)}^{3}}

ⓘ Expand by multiplying each term inside the parentheses:

(8 x + 3) \cdot (16 x + 6)

8 \cdot x \cdot 16 \cdot x

+ 8 x \cdot 6

+ 3 \cdot 16 x

+ 3 \cdot 6

- 32 x \cdot (3 + 4 x)

- (96 x + 128 x^{2})

\frac{8 \cdot x \cdot 16 \cdot x + 8 x \cdot 6 + 3 \cdot 16 x + 3 \cdot 6 - (96 x + 128 x^{2})}{{(3 + 8 x)}^{3}}

ⓘ Multiplications are done before additions and subtractions:

8 \cdot 16

8 \cdot (10 + 6)

8 \cdot 10 + 8 \cdot 6

80 + 48

128

8 \cdot 6

48

3 \cdot 16

3 \cdot (10 + 6)

3 \cdot 10 + 3 \cdot 6

30 + 18

48

3 \cdot 6

18

■Multiplying powers with the same base adds the exponents:

x \cdot x

x^{2}

■A minus sign before a parenthesis changes the sign of each of its terms:

- (96 x + 128 x^{2})

- 96 x - 128 x^{2}

\frac{128 x^{2} + 48 x + 48 x + 18 - 96 x - 128 x^{2}}{{(3 + 8 x)}^{3}}

ⓘ Like terms are grouped by adding their coefficients:

48 x + 48 x - 96 x

0

128 x^{2} - 128 x^{2}

0

\frac{0 + 0 + 18}{{(3 + 8 x)}^{3}}

ⓘ Adding 0 changes nothing:

0 + 0 + 18

18

\frac{18}{{(3 + 8 x)}^{3}}

5) You can then resume the differentiation:

ⅰ

\frac{d}{d x} (\frac{2 x \cdot (3 + 4 x)}{{(3 + 8 x)}^{2}})

\frac{18}{{(3 + 8 x)}^{3}}

\frac{18}{{(3 + 8 x)}^{3}}

▶Derivative:

\frac{18}{{(8 x + 3)}^{3}}