Calculus Examples

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

Step 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 $g (x) = {(\frac{x + 1}{2 x - 1})}^{2}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as ${(\frac{x + 1}{2 x - 1})}^{2}$ .

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

Step 1.2

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

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

Step 1.3

Replace all occurrences of $u_{1}$ with ${(\frac{x + 1}{2 x - 1})}^{2}$ .

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

Step 2

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

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Step 2.1

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

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

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 2$ .

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

Step 2.3

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

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

Step 3

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

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

Step 4

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 $f (x) = x + 1$ and $g (x) = 2 x - 1$ .

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

Step 5

Differentiate.

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Step 5.1

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

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

Step 5.2

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

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

Step 5.3

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

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

Step 5.4

Simplify the expression.

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Step 5.4.1

Add $1$ and $0$ .

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

Step 5.4.2

Multiply $2 x - 1$ by $1$ .

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

Step 5.5

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

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

Step 5.6

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

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

Step 5.7

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

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

Step 5.8

Multiply $2$ by $1$ .

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

Step 5.9

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

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

Step 5.10

Combine fractions.

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Step 5.10.1

Add $2$ and $0$ .

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

Step 5.10.2

Multiply $2$ by $- 1$ .

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

Step 5.10.3

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

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

Step 6

Multiply $2 x - 1$ by ${(2 x - 1)}^{2}$ by adding the exponents.

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Step 6.1

Multiply $2 x - 1$ by ${(2 x - 1)}^{2}$ .

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Step 6.1.1

Raise $2 x - 1$ to the power of $1$ .

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

Step 6.1.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 6.2

Add $1$ and $2$ .

$\frac{1}{{(\frac{x + 1}{2 x - 1})}^{2}} \cdot \frac{2 (x + 1) (2 x - 1 - 2 (x + 1))}{{(2 x - 1)}^{3}}$

Step 7

Simplify.

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Step 7.1

Apply the product rule to $\frac{x + 1}{2 x - 1}$ .

$\frac{1}{\frac{{(x + 1)}^{2}}{{(2 x - 1)}^{2}}} \cdot \frac{2 (x + 1) (2 x - 1 - 2 (x + 1))}{{(2 x - 1)}^{3}}$

Step 7.2

Apply the distributive property.

$\frac{1}{\frac{{(x + 1)}^{2}}{{(2 x - 1)}^{2}}} \cdot \frac{(2 x + 2 \cdot 1) (2 x - 1 - 2 (x + 1))}{{(2 x - 1)}^{3}}$

Step 7.3

Apply the distributive property.

$\frac{1}{\frac{{(x + 1)}^{2}}{{(2 x - 1)}^{2}}} \cdot \frac{(2 x + 2 \cdot 1) (2 x - 1 - 2 x - 2 \cdot 1)}{{(2 x - 1)}^{3}}$

Step 7.4

Combine terms.

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Step 7.4.1

Multiply by the reciprocal of the fraction to divide by $\frac{{(x + 1)}^{2}}{{(2 x - 1)}^{2}}$ .

$1 \frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \frac{(2 x + 2 \cdot 1) (2 x - 1 - 2 x - 2 \cdot 1)}{{(2 x - 1)}^{3}}$

Step 7.4.2

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

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{(2 x + 2 \cdot 1) (2 x - 1 - 2 x - 2 \cdot 1)}{{(2 x - 1)}^{3}}$

Step 7.4.3

Multiply $2$ by $1$ .

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{(2 x + 2) (2 x - 1 - 2 x - 2 \cdot 1)}{{(2 x - 1)}^{3}}$

Step 7.4.4

Multiply $- 2$ by $1$ .

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{(2 x + 2) (2 x - 1 - 2 x - 2)}{{(2 x - 1)}^{3}}$

Step 7.4.5

Subtract $2 x$ from $2 x$ .

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{(2 x + 2) (0 - 1 - 2)}{{(2 x - 1)}^{3}}$

Step 7.4.6

Subtract $1$ from $0$ .

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{(2 x + 2) (- 1 - 2)}{{(2 x - 1)}^{3}}$

Step 7.4.7

Subtract $2$ from $- 1$ .

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{(2 x + 2) \cdot - 3}{{(2 x - 1)}^{3}}$

Step 7.4.8

Move $- 3$ to the left of $2 x + 2$ .

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} \cdot \frac{- 3 \cdot (2 x + 2)}{{(2 x - 1)}^{3}}$

Step 7.4.9

Move the negative in front of the fraction.

$\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}} (- \frac{3 (2 x + 2)}{{(2 x - 1)}^{3}})$

Step 7.4.10

Multiply $\frac{{(2 x - 1)}^{2}}{{(x + 1)}^{2}}$ by $\frac{3 (2 x + 2)}{{(2 x - 1)}^{3}}$ .

$- \frac{{(2 x - 1)}^{2} (3 (2 x + 2))}{{(x + 1)}^{2} {(2 x - 1)}^{3}}$

Step 7.4.11

Move $3$ to the left of ${(2 x - 1)}^{2}$ .

$- \frac{3 \cdot {(2 x - 1)}^{2} (2 x + 2)}{{(x + 1)}^{2} {(2 x - 1)}^{3}}$

Step 7.4.12

Cancel the common factor of ${(2 x - 1)}^{2}$ and ${(2 x - 1)}^{3}$ .

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Step 7.4.12.1

Factor ${(2 x - 1)}^{2}$ out of $3 {(2 x - 1)}^{2} (2 x + 2)$ .

$- \frac{{(2 x - 1)}^{2} (3 (2 x + 2))}{{(x + 1)}^{2} {(2 x - 1)}^{3}}$

Step 7.4.12.2

Cancel the common factors.

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Step 7.4.12.2.1

Factor ${(2 x - 1)}^{2}$ out of ${(x + 1)}^{2} {(2 x - 1)}^{3}$ .

$- \frac{{(2 x - 1)}^{2} (3 (2 x + 2))}{{(2 x - 1)}^{2} ({(x + 1)}^{2} (2 x - 1))}$

Step 7.4.12.2.2

Cancel the common factor.

$- \frac{{(2 x - 1)}^{2} (3 (2 x + 2))}{{(2 x - 1)}^{2} ({(x + 1)}^{2} (2 x - 1))}$

Step 7.4.12.2.3

Rewrite the expression.

$- \frac{3 (2 x + 2)}{{(x + 1)}^{2} (2 x - 1)}$

Step 7.5

Simplify the numerator.

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Step 7.5.1

Factor $2$ out of $2 x + 2$ .

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Step 7.5.1.1

Factor $2$ out of $2 x$ .

$- \frac{3 (2 (x) + 2)}{{(x + 1)}^{2} (2 x - 1)}$

Step 7.5.1.2

Factor $2$ out of $2$ .

$- \frac{3 (2 (x) + 2 (1))}{{(x + 1)}^{2} (2 x - 1)}$

Step 7.5.1.3

Factor $2$ out of $2 (x) + 2 (1)$ .

$- \frac{3 (2 (x + 1))}{{(x + 1)}^{2} (2 x - 1)}$

$- \frac{3 \cdot 2 (x + 1)}{{(x + 1)}^{2} (2 x - 1)}$

Step 7.5.2

Multiply $3$ by $2$ .

$- \frac{6 (x + 1)}{{(x + 1)}^{2} (2 x - 1)}$

Step 7.6

Cancel the common factor of $x + 1$ and ${(x + 1)}^{2}$ .

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Step 7.6.1

Factor $x + 1$ out of $6 (x + 1)$ .

$- \frac{(x + 1) \cdot 6}{{(x + 1)}^{2} (2 x - 1)}$

Step 7.6.2

Cancel the common factors.

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Step 7.6.2.1

Factor $x + 1$ out of ${(x + 1)}^{2} (2 x - 1)$ .

$- \frac{(x + 1) \cdot 6}{(x + 1) ((x + 1) (2 x - 1))}$

Step 7.6.2.2

Cancel the common factor.

$- \frac{(x + 1) \cdot 6}{(x + 1) ((x + 1) (2 x - 1))}$

Step 7.6.2.3

Rewrite the expression.

$- \frac{6}{(x + 1) (2 x - 1)}$