Calculus Examples

$y = \frac{4 x + 3}{5 x + 7}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = 4 x + 3$ and $g (x) = 5 x + 7$ .

$\frac{(5 x + 7) \frac{d}{d x} [4 x + 3] - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $4 x + 3$ with respect to $x$ is $\frac{d}{d x} [4 x] + \frac{d}{d x} [3]$ .

$\frac{(5 x + 7) (\frac{d}{d x} [4 x] + \frac{d}{d x} [3]) - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.2

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

$\frac{(5 x + 7) (4 \frac{d}{d x} [x] + \frac{d}{d x} [3]) - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.3

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

$\frac{(5 x + 7) (4 \cdot 1 + \frac{d}{d x} [3]) - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.4

Multiply $4$ by $1$ .

$\frac{(5 x + 7) (4 + \frac{d}{d x} [3]) - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.5

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$\frac{(5 x + 7) (4 + 0) - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $4$ and $0$ .

$\frac{(5 x + 7) \cdot 4 - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.6.2

Move $4$ to the left of $5 x + 7$ .

$\frac{4 \cdot (5 x + 7) - (4 x + 3) \frac{d}{d x} [5 x + 7]}{{(5 x + 7)}^{2}}$

Step 2.7

By the Sum Rule, the derivative of $5 x + 7$ with respect to $x$ is $\frac{d}{d x} [5 x] + \frac{d}{d x} [7]$ .

$\frac{4 (5 x + 7) - (4 x + 3) (\frac{d}{d x} [5 x] + \frac{d}{d x} [7])}{{(5 x + 7)}^{2}}$

Step 2.8

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$\frac{4 (5 x + 7) - (4 x + 3) (5 \frac{d}{d x} [x] + \frac{d}{d x} [7])}{{(5 x + 7)}^{2}}$

Step 2.9

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

$\frac{4 (5 x + 7) - (4 x + 3) (5 \cdot 1 + \frac{d}{d x} [7])}{{(5 x + 7)}^{2}}$

Step 2.10

Multiply $5$ by $1$ .

$\frac{4 (5 x + 7) - (4 x + 3) (5 + \frac{d}{d x} [7])}{{(5 x + 7)}^{2}}$

Step 2.11

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

$\frac{4 (5 x + 7) - (4 x + 3) (5 + 0)}{{(5 x + 7)}^{2}}$

Step 2.12

Simplify the expression.

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Step 2.12.1

Add $5$ and $0$ .

$\frac{4 (5 x + 7) - (4 x + 3) \cdot 5}{{(5 x + 7)}^{2}}$

Step 2.12.2

Multiply $5$ by $- 1$ .

$\frac{4 (5 x + 7) - 5 (4 x + 3)}{{(5 x + 7)}^{2}}$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\frac{4 (5 x) + 4 \cdot 7 - 5 (4 x + 3)}{{(5 x + 7)}^{2}}$

Step 3.2

Apply the distributive property.

$\frac{4 (5 x) + 4 \cdot 7 - 5 (4 x) - 5 \cdot 3}{{(5 x + 7)}^{2}}$

Step 3.3

Simplify the numerator.

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Step 3.3.1

Simplify each term.

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Step 3.3.1.1

Multiply $5$ by $4$ .

$\frac{20 x + 4 \cdot 7 - 5 (4 x) - 5 \cdot 3}{{(5 x + 7)}^{2}}$

Step 3.3.1.2

Multiply $4$ by $7$ .

$\frac{20 x + 28 - 5 (4 x) - 5 \cdot 3}{{(5 x + 7)}^{2}}$

Step 3.3.1.3

Multiply $4$ by $- 5$ .

$\frac{20 x + 28 - 20 x - 5 \cdot 3}{{(5 x + 7)}^{2}}$

Step 3.3.1.4

Multiply $- 5$ by $3$ .

$\frac{20 x + 28 - 20 x - 15}{{(5 x + 7)}^{2}}$

Step 3.3.2

Combine the opposite terms in $20 x + 28 - 20 x - 15$ .

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Step 3.3.2.1

Subtract $20 x$ from $20 x$ .

$\frac{0 + 28 - 15}{{(5 x + 7)}^{2}}$

Step 3.3.2.2

Add $0$ and $28$ .

$\frac{28 - 15}{{(5 x + 7)}^{2}}$

Step 3.3.3

Subtract $15$ from $28$ .

$\frac{13}{{(5 x + 7)}^{2}}$