Calculus Examples

$y = \frac{\ln (4 x)}{9 x}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} (\frac{\ln (4 x)}{9 x})$

Step 2

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $\frac{1}{9}$ is constant with respect to $x$ , the derivative of $\frac{\ln (4 x)}{9 x}$ with respect to $x$ is $\frac{1}{9} \frac{d}{d x} [\frac{\ln (4 x)}{x}]$ .

$\frac{1}{9} \frac{d}{d x} [\frac{\ln (4 x)}{x}]$

Step 3.2

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 $f (x) = \ln (4 x)$ and $g (x) = x$ .

$\frac{1}{9} \cdot \frac{x \frac{d}{d x} [\ln (4 x)] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.3

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 $g (x) = 4 x$ .

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Step 3.3.1

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

$\frac{1}{9} \cdot \frac{x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [4 x]) - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.3.2

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

$\frac{1}{9} \cdot \frac{x (\frac{1}{u} \frac{d}{d x} [4 x]) - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.3.3

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

$\frac{1}{9} \cdot \frac{x (\frac{1}{4 x} \frac{d}{d x} [4 x]) - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4

Differentiate.

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Step 3.4.1

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

$\frac{1}{9} \cdot \frac{\frac{x}{4 x} \frac{d}{d x} [4 x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.2

Cancel the common factor of $x$ .

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Step 3.4.2.1

Cancel the common factor.

$\frac{1}{9} \cdot \frac{\frac{x}{4 x} \frac{d}{d x} [4 x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.2.2

Rewrite the expression.

$\frac{1}{9} \cdot \frac{\frac{1}{4} \frac{d}{d x} [4 x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.3

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

$\frac{1}{9} \cdot \frac{\frac{1}{4} (4 \frac{d}{d x} [x]) - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.4

Simplify terms.

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Step 3.4.4.1

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

$\frac{1}{9} \cdot \frac{\frac{4}{4} \frac{d}{d x} [x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.4.2

Cancel the common factor of $4$ .

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Step 3.4.4.2.1

Cancel the common factor.

$\frac{1}{9} \cdot \frac{\frac{4}{4} \frac{d}{d x} [x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.4.2.2

Rewrite the expression.

$\frac{1}{9} \cdot \frac{1 \frac{d}{d x} [x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.4.3

Multiply $\frac{d}{d x} [x]$ by $1$ .

$\frac{1}{9} \cdot \frac{\frac{d}{d x} [x] - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.5

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

$\frac{1}{9} \cdot \frac{1 - \ln (4 x) \frac{d}{d x} [x]}{x^{2}}$

Step 3.4.6

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

$\frac{1}{9} \cdot \frac{1 - \ln (4 x) \cdot 1}{x^{2}}$

Step 3.4.7

Combine fractions.

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Step 3.4.7.1

Multiply $- 1$ by $1$ .

$\frac{1}{9} \cdot \frac{1 - \ln (4 x)}{x^{2}}$

Step 3.4.7.2

Multiply $\frac{1}{9}$ by $\frac{1 - \ln (4 x)}{x^{2}}$ .

$\frac{1 - \ln (4 x)}{9 x^{2}}$

Step 4

Reform the equation by setting the left side equal to the right side.

$y' = \frac{1 - \ln (4 x)}{9 x^{2}}$

Step 5

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{1 - \ln (4 x)}{9 x^{2}}$