Calculus Examples

$x \ln (2 x + 1)$

Step 1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = \ln (2 x + 1)$ .

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

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$x (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [2 x + 1]) + \ln (2 x + 1) \frac{d}{d x} [x]$

Step 2.2

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

$x (\frac{1}{u} \frac{d}{d x} [2 x + 1]) + \ln (2 x + 1) \frac{d}{d x} [x]$

Step 2.3

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Multiply $2$ by $1$ .

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

Step 3.6

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

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

Step 3.7

Combine fractions.

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Step 3.7.1

Add $2$ and $0$ .

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

Step 3.7.2

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

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

Step 3.7.3

Move $2$ to the left of $x$ .

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

Step 3.8

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

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

Step 3.9

Multiply $\ln (2 x + 1)$ by $1$ .

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify.

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Step 6.1

Simplify the numerator.

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Step 6.1.1

Simplify each term.

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Step 6.1.1.1

Apply the distributive property.

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

Step 6.1.1.2

Rewrite using the commutative property of multiplication.

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

Step 6.1.1.3

Multiply $\ln (2 x + 1)$ by $1$ .

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

Step 6.1.1.4

Simplify $2 \ln (2 x + 1)$ by moving $2$ inside the logarithm.

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

Step 6.1.2

Reorder factors in $2 x + \ln ({(2 x + 1)}^{2}) x + \ln (2 x + 1)$ .

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

Step 6.2

Reorder terms.

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