Find the Derivative – d/dx natural log of 2x+1

$\ln (2 x + 1)$

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 $g (x) = 2 x + 1$ .

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

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

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

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

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

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

Differentiate.

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

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

Combine fractions.

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Add $2$ and $0$ .

$\frac{1}{2 x + 1} \cdot 2$

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

$\frac{2}{2 x + 1}$

Find the Derivative – d/dx natural log of 2x+1

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