derivative of y=(7ln(x))/(3x+4) — step-by-step solution

We save the definition:

y = \frac{7 \cdot \ln (x)}{3 x + 4}

1) Let's differentiate:

\frac{d}{d x} (\frac{7 \cdot \ln (x)}{3 x + 4})

For

p1

7 \cdot \ln (x)

and

q2

3 x + 4

, 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 differentiate

p1

\frac{d}{d x} (7 \cdot \ln (x))

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

\frac{d}{d x} (7 \cdot \ln (x))

7 \cdot \frac{d}{d x} (\ln (x))

■The derivative of ln(x) is 1/x:

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

\frac{1}{x}

7 \cdot \frac{1}{x}

ⓘ Multiplying a fraction by a number multiplies its numerator:

7 \cdot \frac{1}{x}

\frac{7 \cdot 1}{x}

\frac{7}{x}

\frac{7}{x}

3) Let's differentiate

q2

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

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

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

\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{\frac{7}{x} \cdot (3 x + 4) - 7 \cdot \ln (x) \cdot 3}{{(3 x + 4)}^{2}}

ⓘ Multiplying a fraction by a term multiplies its numerator:

\frac{7}{x} \cdot (3 x + 4)

\frac{7 \cdot (3 x + 4)}{x}

■Multiplications are done before additions and subtractions:

7 \cdot 3

21

■To add fractions, put them over the same denominator, then add the numerators:

\frac{7 \cdot (3 x + 4)}{x} - 21 \cdot \ln (x)

\frac{7 \cdot (3 x + 4) - 21 \cdot \ln (x) \cdot x}{x}

\frac{\frac{7 \cdot (3 x + 4) - 21 \cdot \ln (x) \cdot x}{x}}{{(3 x + 4)}^{2}}

ⓘ Dividing a fraction by

{(3 x + 4)}^{2}

is multiplying its denominator by

{(3 x + 4)}^{2}

\frac{\frac{7 \cdot (3 x + 4) - 21 \cdot \ln (x) \cdot x}{x}}{{(3 x + 4)}^{2}}

\frac{7 \cdot (3 x + 4) - 21 \cdot \ln (x) \cdot x}{x \cdot {(3 x + 4)}^{2}}

\frac{7 \cdot (3 x + 4) - 21 \cdot \ln (x) \cdot x}{x \cdot {(3 x + 4)}^{2}}

ⓘ Expand by multiplying each term inside the parentheses:

7 \cdot (3 x + 4)

21 x + 28

\frac{21 x + 28 - 21 \cdot \ln (x) \cdot x}{x \cdot {(3 x + 4)}^{2}}

5) You can then resume the derivation:

ⅰ

\frac{d}{d x} (\frac{7 \cdot \ln (x)}{3 x + 4})

\frac{21 x + 28 - 21 \cdot \ln (x) \cdot x}{x \cdot {(3 x + 4)}^{2}}

\frac{21 x + 28 - 21 \cdot \ln (x) \cdot x}{x \cdot {(3 x + 4)}^{2}}

▶Derivative:

\frac{- 21 x \cdot \ln (x) + 21 x + 28}{x \cdot {(3 x + 4)}^{2}}