Calculus Examples

$p = \frac{5 q + 8}{6 q + 1}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d p} [\frac{f (p)}{g (p)}]$ is $\frac{g (p) \frac{d}{d p} [f (p)] - f (p) \frac{d}{d p} [g (p)]}{{g (p)}^{2}}$ where $f (p) = 5 q + 8$ and $g (p) = 6 q + 1$ .

$\frac{(6 q + 1) \frac{d}{d p} [5 q + 8] - (5 q + 8) \frac{d}{d p} [6 q + 1]}{{(6 q + 1)}^{2}}$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $5 q + 8$ with respect to $p$ is $\frac{d}{d p} [5 q] + \frac{d}{d p} [8]$ .

$\frac{(6 q + 1) (\frac{d}{d p} [5 q] + \frac{d}{d p} [8]) - (5 q + 8) \frac{d}{d p} [6 q + 1]}{{(6 q + 1)}^{2}}$

Step 2.2

Since $5 q$ is constant with respect to $p$ , the derivative of $5 q$ with respect to $p$ is $0$ .

$\frac{(6 q + 1) (0 + \frac{d}{d p} [8]) - (5 q + 8) \frac{d}{d p} [6 q + 1]}{{(6 q + 1)}^{2}}$

Step 2.3

Since $8$ is constant with respect to $p$ , the derivative of $8$ with respect to $p$ is $0$ .

$\frac{(6 q + 1) (0 + 0) - (5 q + 8) \frac{d}{d p} [6 q + 1]}{{(6 q + 1)}^{2}}$

Step 2.4

Add $0$ and $0$ .

$\frac{(6 q + 1) \cdot 0 - (5 q + 8) \frac{d}{d p} [6 q + 1]}{{(6 q + 1)}^{2}}$

Step 2.5

By the Sum Rule, the derivative of $6 q + 1$ with respect to $p$ is $\frac{d}{d p} [6 q] + \frac{d}{d p} [1]$ .

$\frac{(6 q + 1) \cdot 0 - (5 q + 8) (\frac{d}{d p} [6 q] + \frac{d}{d p} [1])}{{(6 q + 1)}^{2}}$

Step 2.6

Since $6 q$ is constant with respect to $p$ , the derivative of $6 q$ with respect to $p$ is $0$ .

$\frac{(6 q + 1) \cdot 0 - (5 q + 8) (0 + \frac{d}{d p} [1])}{{(6 q + 1)}^{2}}$

Step 2.7

Since $1$ is constant with respect to $p$ , the derivative of $1$ with respect to $p$ is $0$ .

$\frac{(6 q + 1) \cdot 0 - (5 q + 8) (0 + 0)}{{(6 q + 1)}^{2}}$

Step 2.8

Add $0$ and $0$ .

$\frac{(6 q + 1) \cdot 0 - (5 q + 8) \cdot 0}{{(6 q + 1)}^{2}}$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\frac{6 q \cdot 0 + 1 \cdot 0 - (5 q + 8) \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.2

Apply the distributive property.

$\frac{6 q \cdot 0 + 1 \cdot 0 + (- (5 q) - 1 \cdot 8) \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.3

Apply the distributive property.

$\frac{6 q \cdot 0 + 1 \cdot 0 - (5 q) \cdot 0 - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4

Simplify the numerator.

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Step 3.4.1

Simplify each term.

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Step 3.4.1.1

Multiply $6 q \cdot 0$ .

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Step 3.4.1.1.1

Multiply $0$ by $6$ .

$\frac{0 q + 1 \cdot 0 - (5 q) \cdot 0 - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.1.2

Multiply $0$ by $q$ .

$\frac{0 + 1 \cdot 0 - (5 q) \cdot 0 - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.2

Multiply $0$ by $1$ .

$\frac{0 + 0 - (5 q) \cdot 0 - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.3

Multiply $5$ by $- 1$ .

$\frac{0 + 0 - 5 q \cdot 0 - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.4

Multiply $- 5 q \cdot 0$ .

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Step 3.4.1.4.1

Multiply $0$ by $- 5$ .

$\frac{0 + 0 + 0 q - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.4.2

Multiply $0$ by $q$ .

$\frac{0 + 0 + 0 - 1 \cdot 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.5

Multiply $- 1 \cdot 8 \cdot 0$ .

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Step 3.4.1.5.1

Multiply $- 1$ by $8$ .

$\frac{0 + 0 + 0 - 8 \cdot 0}{{(6 q + 1)}^{2}}$

Step 3.4.1.5.2

Multiply $- 8$ by $0$ .

$\frac{0 + 0 + 0 + 0}{{(6 q + 1)}^{2}}$

Step 3.4.2

Add $0$ and $0$ .

$\frac{0 + 0 + 0}{{(6 q + 1)}^{2}}$

Step 3.4.3

Add $0$ and $0$ .

$\frac{0 + 0}{{(6 q + 1)}^{2}}$

Step 3.4.4

Add $0$ and $0$ .

$\frac{0}{{(6 q + 1)}^{2}}$

Step 3.5

Divide $0$ by ${(6 q + 1)}^{2}$ .

$0$