Calculus Examples

$f (x) = \ln (x + 5) + \ln (2 x - 1) + \ln (4 - x)$

Step 1

By the Sum Rule, the derivative of $\ln (x + 5) + \ln (2 x - 1) + \ln (4 - x)$ with respect to $x$ is $\frac{d}{d x} [\ln (x + 5)] + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$ .

$\frac{d}{d x} [\ln (x + 5)] + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2

Evaluate $\frac{d}{d x} [\ln (x + 5)]$ .

Tap for more steps...

Step 2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x + 5$ .

Tap for more steps...

Step 2.1.1

To apply the Chain Rule, set $u_{1}$ as $x + 5$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [x + 5] + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2.1.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

$\frac{1}{u_{1}} \frac{d}{d x} [x + 5] + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2.1.3

Replace all occurrences of $u_{1}$ with $x + 5$ .

$\frac{1}{x + 5} \frac{d}{d x} [x + 5] + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2.2

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

$\frac{1}{x + 5} (\frac{d}{d x} [x] + \frac{d}{d x} [5]) + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2.3

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

$\frac{1}{x + 5} (1 + \frac{d}{d x} [5]) + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2.4

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

$\frac{1}{x + 5} (1 + 0) + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 2.5

Add $1$ and $0$ .

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

Step 2.6

Multiply $\frac{1}{x + 5}$ by $1$ .

$\frac{1}{x + 5} + \frac{d}{d x} [\ln (2 x - 1)] + \frac{d}{d x} [\ln (4 - x)]$

Step 3

Evaluate $\frac{d}{d x} [\ln (2 x - 1)]$ .

Tap for more steps...

Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 2 x - 1$ .

Tap for more steps...

Step 3.1.1

To apply the Chain Rule, set $u_{2}$ as $2 x - 1$ .

$\frac{1}{x + 5} + \frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [2 x - 1] + \frac{d}{d x} [\ln (4 - x)]$

Step 3.1.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{1}{x + 5} + \frac{1}{u_{2}} \frac{d}{d x} [2 x - 1] + \frac{d}{d x} [\ln (4 - x)]$

Step 3.1.3

Replace all occurrences of $u_{2}$ with $2 x - 1$ .

$\frac{1}{x + 5} + \frac{1}{2 x - 1} \frac{d}{d x} [2 x - 1] + \frac{d}{d x} [\ln (4 - x)]$

Step 3.2

By the Sum Rule, the derivative of $2 x - 1$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]$ .

$\frac{1}{x + 5} + \frac{1}{2 x - 1} (\frac{d}{d x} [2 x] + \frac{d}{d x} [- 1]) + \frac{d}{d x} [\ln (4 - x)]$

Step 3.3

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

$\frac{1}{x + 5} + \frac{1}{2 x - 1} (2 \frac{d}{d x} [x] + \frac{d}{d x} [- 1]) + \frac{d}{d x} [\ln (4 - x)]$

Step 3.4

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

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

Step 3.5

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

$\frac{1}{x + 5} + \frac{1}{2 x - 1} (2 \cdot 1 + 0) + \frac{d}{d x} [\ln (4 - x)]$

Step 3.6

Multiply $2$ by $1$ .

$\frac{1}{x + 5} + \frac{1}{2 x - 1} (2 + 0) + \frac{d}{d x} [\ln (4 - x)]$

Step 3.7

Add $2$ and $0$ .

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

Step 3.8

Combine $\frac{1}{2 x - 1}$ and $2$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{d}{d x} [\ln (4 - x)]$

Step 4

Evaluate $\frac{d}{d x} [\ln (4 - x)]$ .

Tap for more steps...

Step 4.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = 4 - x$ .

Tap for more steps...

Step 4.1.1

To apply the Chain Rule, set $u_{3}$ as $4 - x$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{d}{d u_{3}} [\ln (u_{3})] \frac{d}{d x} [4 - x]$

Step 4.1.2

The derivative of $\ln (u_{3})$ with respect to $u_{3}$ is $\frac{1}{u_{3}}$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{u_{3}} \frac{d}{d x} [4 - x]$

Step 4.1.3

Replace all occurrences of $u_{3}$ with $4 - x$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{4 - x} \frac{d}{d x} [4 - x]$

Step 4.2

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

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

Step 4.3

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{4 - x} (0 + \frac{d}{d x} [- x])$

Step 4.4

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{4 - x} (0 - \frac{d}{d x} [x])$

Step 4.5

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

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{4 - x} (0 - 1 \cdot 1)$

Step 4.6

Multiply $- 1$ by $1$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{4 - x} (0 - 1)$

Step 4.7

Subtract $1$ from $0$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{1}{4 - x} \cdot - 1$

Step 4.8

Combine $\frac{1}{4 - x}$ and $- 1$ .

$\frac{1}{x + 5} + \frac{2}{2 x - 1} + \frac{- 1}{4 - x}$

Step 4.9

Move the negative in front of the fraction.

$\frac{1}{x + 5} + \frac{2}{2 x - 1} - \frac{1}{4 - x}$

Step 5

Combine terms.

Tap for more steps...

Step 5.1

To write $\frac{1}{x + 5}$ as a fraction with a common denominator, multiply by $\frac{2 x - 1}{2 x - 1}$ .

$\frac{1}{x + 5} \cdot \frac{2 x - 1}{2 x - 1} + \frac{2}{2 x - 1} - \frac{1}{4 - x}$

Step 5.2

To write $\frac{2}{2 x - 1}$ as a fraction with a common denominator, multiply by $\frac{x + 5}{x + 5}$ .

$\frac{1}{x + 5} \cdot \frac{2 x - 1}{2 x - 1} + \frac{2}{2 x - 1} \cdot \frac{x + 5}{x + 5} - \frac{1}{4 - x}$

Step 5.3

Write each expression with a common denominator of $(x + 5) (2 x - 1)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 5.3.1

Multiply $\frac{1}{x + 5}$ by $\frac{2 x - 1}{2 x - 1}$ .

$\frac{2 x - 1}{(x + 5) (2 x - 1)} + \frac{2}{2 x - 1} \cdot \frac{x + 5}{x + 5} - \frac{1}{4 - x}$

Step 5.3.2

Multiply $\frac{2}{2 x - 1}$ by $\frac{x + 5}{x + 5}$ .

$\frac{2 x - 1}{(x + 5) (2 x - 1)} + \frac{2 (x + 5)}{(2 x - 1) (x + 5)} - \frac{1}{4 - x}$

Step 5.3.3

Reorder the factors of $(x + 5) (2 x - 1)$ .

$\frac{2 x - 1}{(2 x - 1) (x + 5)} + \frac{2 (x + 5)}{(2 x - 1) (x + 5)} - \frac{1}{4 - x}$

Step 5.4

Combine the numerators over the common denominator.

$\frac{2 x - 1 + 2 (x + 5)}{(2 x - 1) (x + 5)} - \frac{1}{4 - x}$

Step 5.5

To write $\frac{2 x - 1 + 2 (x + 5)}{(2 x - 1) (x + 5)}$ as a fraction with a common denominator, multiply by $\frac{4 - x}{4 - x}$ .

$\frac{2 x - 1 + 2 (x + 5)}{(2 x - 1) (x + 5)} \cdot \frac{4 - x}{4 - x} - \frac{1}{4 - x}$

Step 5.6

To write $- \frac{1}{4 - x}$ as a fraction with a common denominator, multiply by $\frac{(2 x - 1) (x + 5)}{(2 x - 1) (x + 5)}$ .

$\frac{2 x - 1 + 2 (x + 5)}{(2 x - 1) (x + 5)} \cdot \frac{4 - x}{4 - x} - \frac{1}{4 - x} \cdot \frac{(2 x - 1) (x + 5)}{(2 x - 1) (x + 5)}$

Step 5.7

Write each expression with a common denominator of $(2 x - 1) (x + 5) (4 - x)$ , by multiplying each by an appropriate factor of $1$ .

Tap for more steps...

Step 5.7.1

Multiply $\frac{2 x - 1 + 2 (x + 5)}{(2 x - 1) (x + 5)}$ by $\frac{4 - x}{4 - x}$ .

$\frac{(2 x - 1 + 2 (x + 5)) (4 - x)}{(2 x - 1) (x + 5) (4 - x)} - \frac{1}{4 - x} \cdot \frac{(2 x - 1) (x + 5)}{(2 x - 1) (x + 5)}$

Step 5.7.2

Multiply $\frac{1}{4 - x}$ by $\frac{(2 x - 1) (x + 5)}{(2 x - 1) (x + 5)}$ .

$\frac{(2 x - 1 + 2 (x + 5)) (4 - x)}{(2 x - 1) (x + 5) (4 - x)} - \frac{(2 x - 1) (x + 5)}{(4 - x) ((2 x - 1) (x + 5))}$

Step 5.7.3

Reorder the factors of $(2 x - 1) (x + 5) (4 - x)$ .

$\frac{(2 x - 1 + 2 (x + 5)) (4 - x)}{(2 x - 1) (4 - x) (x + 5)} - \frac{(2 x - 1) (x + 5)}{(4 - x) ((2 x - 1) (x + 5))}$

Step 5.7.4

Reorder the factors of $(4 - x) ((2 x - 1) (x + 5))$ .

$\frac{(2 x - 1 + 2 (x + 5)) (4 - x)}{(2 x - 1) (4 - x) (x + 5)} - \frac{(2 x - 1) (x + 5)}{(2 x - 1) (4 - x) (x + 5)}$

Step 5.8

Combine the numerators over the common denominator.

$\frac{(2 x - 1 + 2 (x + 5)) (4 - x) - (2 x - 1) (x + 5)}{(2 x - 1) (4 - x) (x + 5)}$