Find the Derivative Using Quotient Rule - d/db (b^3+4b^2-3b+126)/(b+7)

$\frac{b^{3} + 4 b^{2} - 3 b + 126}{b + 7}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d b} [\frac{f (b)}{g (b)}]$ is $\frac{g (b) \frac{d}{d b} [f (b)] - f (b) \frac{d}{d b} [g (b)]}{{g (b)}^{2}}$ where $f (b) = b^{3} + 4 b^{2} - 3 b + 126$ and $g (b) = b + 7$ .

$\frac{(b + 7) \frac{d}{d b} [b^{3} + 4 b^{2} - 3 b + 126] - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Step 2

Differentiate.

By the Sum Rule, the derivative of $b^{3} + 4 b^{2} - 3 b + 126$ with respect to $b$ is $\frac{d}{d b} [b^{3}] + \frac{d}{d b} [4 b^{2}] + \frac{d}{d b} [- 3 b] + \frac{d}{d b} [126]$ .

$\frac{(b + 7) (\frac{d}{d b} [b^{3}] + \frac{d}{d b} [4 b^{2}] + \frac{d}{d b} [- 3 b] + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Differentiate using the Power Rule which states that $\frac{d}{d b} [b^{n}]$ is $n b^{n - 1}$ where $n = 3$ .

$\frac{(b + 7) (3 b^{2} + \frac{d}{d b} [4 b^{2}] + \frac{d}{d b} [- 3 b] + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Step 3

Evaluate $\frac{d}{d b} [4 b^{2}]$ .

Since $4$ is constant with respect to $b$ , the derivative of $4 b^{2}$ with respect to $b$ is $4 \frac{d}{d b} [b^{2}]$ .

$\frac{(b + 7) (3 b^{2} + 4 \frac{d}{d b} [b^{2}] + \frac{d}{d b} [- 3 b] + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Differentiate using the Power Rule which states that $\frac{d}{d b} [b^{n}]$ is $n b^{n - 1}$ where $n = 2$ .

$\frac{(b + 7) (3 b^{2} + 4 (2 b) + \frac{d}{d b} [- 3 b] + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Multiply $2$ by $4$ .

$\frac{(b + 7) (3 b^{2} + 8 b + \frac{d}{d b} [- 3 b] + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Step 4

Evaluate $\frac{d}{d b} [- 3 b]$ .

Since $- 3$ is constant with respect to $b$ , the derivative of $- 3 b$ with respect to $b$ is $- 3 \frac{d}{d b} [b]$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3 \frac{d}{d b} [b] + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Differentiate using the Power Rule which states that $\frac{d}{d b} [b^{n}]$ is $n b^{n - 1}$ where $n = 1$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3 \cdot 1 + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Multiply $- 3$ by $1$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3 + \frac{d}{d b} [126]) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Step 5

Differentiate.

Since $126$ is constant with respect to $b$ , the derivative of $126$ with respect to $b$ is $0$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3 + 0) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

Add $3 b^{2} + 8 b - 3$ and $0$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3) - (b^{3} + 4 b^{2} - 3 b + 126) \frac{d}{d b} [b + 7]}{{(b + 7)}^{2}}$

By the Sum Rule, the derivative of $b + 7$ with respect to $b$ is $\frac{d}{d b} [b] + \frac{d}{d b} [7]$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3) - (b^{3} + 4 b^{2} - 3 b + 126) (\frac{d}{d b} [b] + \frac{d}{d b} [7])}{{(b + 7)}^{2}}$

Differentiate using the Power Rule which states that $\frac{d}{d b} [b^{n}]$ is $n b^{n - 1}$ where $n = 1$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3) - (b^{3} + 4 b^{2} - 3 b + 126) (1 + \frac{d}{d b} [7])}{{(b + 7)}^{2}}$

Since $7$ is constant with respect to $b$ , the derivative of $7$ with respect to $b$ is $0$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3) - (b^{3} + 4 b^{2} - 3 b + 126) (1 + 0)}{{(b + 7)}^{2}}$

Add $1$ and $0$ .

$\frac{(b + 7) (3 b^{2} + 8 b - 3) - (b^{3} + 4 b^{2} - 3 b + 126) \cdot 1}{{(b + 7)}^{2}}$

Step 6

Simplify.

Apply the distributive property.

$\frac{(b + 7) (3 b^{2} + 8 b - 3) + (- b^{3} - (4 b^{2}) - (- 3 b) - 1 \cdot 126) \cdot 1}{{(b + 7)}^{2}}$

Apply the distributive property.

$\frac{(b + 7) (3 b^{2} + 8 b - 3) - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Simplify the numerator.

Simplify each term.

Expand $(b + 7) (3 b^{2} + 8 b - 3)$ by multiplying each term in the first expression by each term in the second expression.

$\frac{b (3 b^{2}) + b (8 b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Simplify each term.

Rewrite using the commutative property of multiplication.

$\frac{3 b \cdot b^{2} + b (8 b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $b$ by $b^{2}$ by adding the exponents.

Move $b^{2}$ .

$\frac{3 (b^{2} b) + b (8 b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $b^{2}$ by $b$ .

Raise $b$ to the power of $1$ .

$\frac{3 (b^{2} b^{1}) + b (8 b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{3 b^{2 + 1} + b (8 b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Add $2$ and $1$ .

$\frac{3 b^{3} + b (8 b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Rewrite using the commutative property of multiplication.

$\frac{3 b^{3} + 8 b \cdot b + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $b$ by $b$ by adding the exponents.

Move $b$ .

$\frac{3 b^{3} + 8 (b \cdot b) + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $b$ by $b$ .

$\frac{3 b^{3} + 8 b^{2} + b \cdot - 3 + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Move $- 3$ to the left of $b$ .

$\frac{3 b^{3} + 8 b^{2} - 3 \cdot b + 7 (3 b^{2}) + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $3$ by $7$ .

$\frac{3 b^{3} + 8 b^{2} - 3 b + 21 b^{2} + 7 (8 b) + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $8$ by $7$ .

$\frac{3 b^{3} + 8 b^{2} - 3 b + 21 b^{2} + 56 b + 7 \cdot - 3 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $7$ by $- 3$ .

$\frac{3 b^{3} + 8 b^{2} - 3 b + 21 b^{2} + 56 b - 21 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Add $8 b^{2}$ and $21 b^{2}$ .

$\frac{3 b^{3} + 29 b^{2} - 3 b + 56 b - 21 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Add $- 3 b$ and $56 b$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} \cdot 1 - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $- 1$ by $1$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - (4 b^{2}) \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $4$ by $- 1$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - 4 b^{2} \cdot 1 - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $- 4$ by $1$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - 4 b^{2} - (- 3 b) \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $- 3$ by $- 1$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - 4 b^{2} + 3 b \cdot 1 - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $3$ by $1$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - 4 b^{2} + 3 b - 1 \cdot 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $- 1 \cdot 126 \cdot 1$ .

Multiply $- 1$ by $126$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - 4 b^{2} + 3 b - 126 \cdot 1}{{(b + 7)}^{2}}$

Multiply $- 126$ by $1$ .

$\frac{3 b^{3} + 29 b^{2} + 53 b - 21 - b^{3} - 4 b^{2} + 3 b - 126}{{(b + 7)}^{2}}$

Subtract $b^{3}$ from $3 b^{3}$ .

$\frac{2 b^{3} + 29 b^{2} + 53 b - 21 - 4 b^{2} + 3 b - 126}{{(b + 7)}^{2}}$

Subtract $4 b^{2}$ from $29 b^{2}$ .

$\frac{2 b^{3} + 25 b^{2} + 53 b - 21 + 3 b - 126}{{(b + 7)}^{2}}$

Add $53 b$ and $3 b$ .

$\frac{2 b^{3} + 25 b^{2} + 56 b - 21 - 126}{{(b + 7)}^{2}}$

Subtract $126$ from $- 21$ .

$\frac{2 b^{3} + 25 b^{2} + 56 b - 147}{{(b + 7)}^{2}}$

Find the Derivative Using Quotient Rule - d/db (b^3+4b^2-3b+126)/(b+7)

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