derivative of h(x)=(x^2+4x+1)/(3x+5)

We save the definition:

h (x) = \frac{x^{2} + 4 x + 1}{3 x + 5}

1) Let's perform the differentiation:

\frac{d}{d x} (\frac{x^{2} + 4 x + 1}{3 x + 5})

For

p1

x^{2} + 4 x + 1

and

q2

3 x + 5

, use the differentiation rule:

\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 perform the differentiation

p1

\frac{d}{d x} (x^{2} + 4 x + 1)

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

\frac{d}{d x} (x^{2} + 4 x + 1)

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (4 x) + 0

\frac{d}{d x} (x^{2}) + \frac{d}{d x} (4 x) + 0

ⓘ The derivative of x^n is n*x^(n-1):

\frac{d}{d x} (x^{2})

2 x

■Adding 0 changes nothing:

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

\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

2 x + 4

3) Let's perform the differentiation

q2

\frac{d}{d x} (3 x + 5)

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

\frac{d}{d x} (3 x + 5)

\frac{d}{d x} (3 x) + 0

\frac{d}{d x} (3 x) + 0

ⓘ Adding 0 changes nothing:

\frac{d}{d x} (3 x) + 0

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

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

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

3

3

4) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

\frac{(2 x + 4) \cdot (3 x + 5) - (x^{2} + 4 x + 1) \cdot 3}{{(3 x + 5)}^{2}}

ⓘ Expand by multiplying each term inside the parentheses:

(2 x + 4) \cdot (3 x + 5)

2 \cdot x \cdot 3 \cdot x

+ 2 x \cdot 5

+ 4 \cdot 3 x

+ 4 \cdot 5

- (x^{2} + 4 x + 1) \cdot 3

- (x^{2} \cdot 3 + 12 x + 3)

\frac{2 \cdot x \cdot 3 \cdot x + 2 x \cdot 5 + 4 \cdot 3 x + 4 \cdot 5 - (x^{2} \cdot 3 + 12 x + 3)}{{(3 x + 5)}^{2}}

ⓘ Multiplications are done before additions and subtractions:

2 \cdot 3

6

2 \cdot 5

10

4 \cdot 3

12

4 \cdot 5

20

■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:

- (x^{2} \cdot 3 + 12 x + 3)

- x^{2} \cdot 3 - 12 x - 3

\frac{6 x^{2} + 10 x + 12 x + 20 - x^{2} \cdot 3 - 12 x - 3}{{(3 x + 5)}^{2}}

ⓘ The numerator and the denominator are computed before the division:

20 - 3

17

■Like terms are grouped by adding their coefficients:

12 x

+ 17

- x^{2} \cdot 3

- 12 x

17 - x^{2} \cdot 3

6 x^{2} - x^{2} \cdot 3

3 x^{2}

\frac{3 x^{2} + 10 x + 17}{{(3 x + 5)}^{2}}

5) You can then resume the differentiation:

ⅰ

\frac{d}{d x} (\frac{x^{2} + 4 x + 1}{3 x + 5})

\frac{3 x^{2} + 10 x + 17}{{(3 x + 5)}^{2}}

▶Derivative:

\frac{3 x^{2} + 10 x + 17}{{(3 x + 5)}^{2}}