Calculus Examples

$g (x) = \frac{x}{x - 6 + 1.5}$

Step 1

Add $- 6$ and $1.5$ .

$\frac{d}{d x} [\frac{x}{x - 4.5}]$

Step 2

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 $f (x) = x$ and $g (x) = x - 4.5$ .

$\frac{(x - 4.5) \frac{d}{d x} [x] - x \frac{d}{d x} [x - 4.5]}{{(x - 4.5)}^{2}}$

Step 3

Differentiate.

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Step 3.1

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

$\frac{(x - 4.5) \cdot 1 - x \frac{d}{d x} [x - 4.5]}{{(x - 4.5)}^{2}}$

Step 3.2

Multiply $x - 4.5$ by $1$ .

$\frac{x - 4.5 - x \frac{d}{d x} [x - 4.5]}{{(x - 4.5)}^{2}}$

Step 3.3

By the Sum Rule, the derivative of $x - 4.5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [- 4.5]$ .

$\frac{x - 4.5 - x (\frac{d}{d x} [x] + \frac{d}{d x} [- 4.5])}{{(x - 4.5)}^{2}}$

Step 3.4

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

$\frac{x - 4.5 - x (1 + \frac{d}{d x} [- 4.5])}{{(x - 4.5)}^{2}}$

Step 3.5

Since $- 4.5$ is constant with respect to $x$ , the derivative of $- 4.5$ with respect to $x$ is $0$ .

$\frac{x - 4.5 - x (1 + 0)}{{(x - 4.5)}^{2}}$

Step 3.6

Simplify by adding terms.

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Step 3.6.1

Add $1$ and $0$ .

$\frac{x - 4.5 - x \cdot 1}{{(x - 4.5)}^{2}}$

Step 3.6.2

Multiply $- 1$ by $1$ .

$\frac{x - 4.5 - x}{{(x - 4.5)}^{2}}$

Step 3.6.3

Subtract $x$ from $x$ .

$\frac{0 - 4.5}{{(x - 4.5)}^{2}}$

Step 3.6.4

Simplify the expression.

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Step 3.6.4.1

Subtract $4.5$ from $0$ .

$\frac{- 4.5}{{(x - 4.5)}^{2}}$

Step 3.6.4.2

Move the negative in front of the fraction.

$- \frac{4.5}{{(x - 4.5)}^{2}}$