Calculus Examples

$y = \ln (4 x - 1) - 3 \ln (x)$

Step 1

By the Sum Rule, the derivative of $\ln (4 x - 1) - 3 \ln (x)$ with respect to $x$ is $\frac{d}{d x} [\ln (4 x - 1)] + \frac{d}{d x} [- 3 \ln (x)]$ .

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

Step 2

Evaluate $\frac{d}{d x} [\ln (4 x - 1)]$ .

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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 $g (x) = 4 x - 1$ .

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Step 2.1.1

To apply the Chain Rule, set $u$ as $4 x - 1$ .

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

Step 2.1.2

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

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

Step 2.1.3

Replace all occurrences of $u$ with $4 x - 1$ .

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

$\frac{1}{4 x - 1} (4 \cdot 1 + 0) + \frac{d}{d x} [- 3 \ln (x)]$

Step 2.6

Multiply $4$ by $1$ .

$\frac{1}{4 x - 1} (4 + 0) + \frac{d}{d x} [- 3 \ln (x)]$

Step 2.7

Add $4$ and $0$ .

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

Step 2.8

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

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

Step 3

Evaluate $\frac{d}{d x} [- 3 \ln (x)]$ .

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Step 3.1

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

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

Step 3.2

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

$\frac{4}{4 x - 1} - 3 \frac{1}{x}$

Step 3.3

Combine $- 3$ and $\frac{1}{x}$ .

$\frac{4}{4 x - 1} + \frac{- 3}{x}$

Step 3.4

Move the negative in front of the fraction.

$\frac{4}{4 x - 1} - \frac{3}{x}$

Step 4

Combine terms.

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Step 4.1

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

$\frac{4}{4 x - 1} \cdot \frac{x}{x} - \frac{3}{x}$

Step 4.2

To write $- \frac{3}{x}$ as a fraction with a common denominator, multiply by $\frac{4 x - 1}{4 x - 1}$ .

$\frac{4}{4 x - 1} \cdot \frac{x}{x} - \frac{3}{x} \cdot \frac{4 x - 1}{4 x - 1}$

Step 4.3

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

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Step 4.3.1

Multiply $\frac{4}{4 x - 1}$ by $\frac{x}{x}$ .

$\frac{4 x}{(4 x - 1) x} - \frac{3}{x} \cdot \frac{4 x - 1}{4 x - 1}$

Step 4.3.2

Multiply $\frac{3}{x}$ by $\frac{4 x - 1}{4 x - 1}$ .

$\frac{4 x}{(4 x - 1) x} - \frac{3 (4 x - 1)}{x (4 x - 1)}$

Step 4.3.3

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

$\frac{4 x}{x (4 x - 1)} - \frac{3 (4 x - 1)}{x (4 x - 1)}$

Step 4.4

Combine the numerators over the common denominator.

$\frac{4 x - 3 (4 x - 1)}{x (4 x - 1)}$