derivative of (4-3x/(3x^2+x)) — step-by-step solution

1) Let's perform the differentiation:

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

2) Let's factor in

ℝ

3 x^{2} + x

You can write the expression in factored form:

x \cdot (3 x + 1)

ⓘ The original denominator must remain nonzero. The fraction is simplified by dividing the numerator and the denominator by their common factor:

\frac{-3 x}{3 x^{2} + x}

- \frac{3}{3 x + 1}

\frac{d}{d x} (4 - \frac{3}{3 x + 1})

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

\frac{d}{d x} (4 - \frac{3}{3 x + 1})

0 + \frac{d}{d x} (- \frac{3}{3 x + 1})

0 + \frac{d}{d x} (- \frac{3}{3 x + 1})

We differentiate

p1

3

and

q2

3 x + 1

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}}

3) Let's differentiate

p1

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

ⓘ The derivative of a constant is zero:

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

0

0

4) Let's perform the differentiation

q2

\frac{d}{d x} (3 x + 1)

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

\frac{d}{d x} (3 x + 1)

\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

5) We evaluate the resulting derivative (

\frac{p1}{q2}

)':

- \frac{0 \cdot (3 x + 1) - 3 \cdot 3}{{(3 x + 1)}^{2}}

ⓘ A product with a zero factor is zero:

0 \cdot (3 x + 1)

0

■Multiplications are done before additions and subtractions:

3 \cdot 3

9

- \frac{0 - 9}{{(3 x + 1)}^{2}}

ⓘ Adding 0 changes nothing:

0 - 9

-9

■Two minus signs cancel out:

- (\frac{-9}{{(3 x + 1)}^{2}})

\frac{9}{{(3 x + 1)}^{2}}

\frac{9}{{(3 x + 1)}^{2}}

6) Then you can continue the differentiation:

ⅰ

\frac{d}{d x} (- \frac{3}{3 x + 1})

\frac{9}{{(3 x + 1)}^{2}}

▶Derivative:

\frac{9}{{(3 x + 1)}^{2}}