Find the Derivative f(x) = natural log of x+5

$f (x) = \ln (x + 5)$

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 $g (x) = x + 5$ .

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To apply the Chain Rule, set $u$ as $x + 5$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x + 5]$

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

$\frac{1}{u} \frac{d}{d x} [x + 5]$

Replace all occurrences of $u$ with $x + 5$ .

$\frac{1}{x + 5} \frac{d}{d x} [x + 5]$

Differentiate.

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By the Sum Rule, the derivative of $x + 5$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [5]$ .

$\frac{1}{x + 5} (\frac{d}{d x} [x] + \frac{d}{d x} [5])$

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

$\frac{1}{x + 5} (1 + \frac{d}{d x} [5])$

Since $5$ is constant with respect to $x$ , the derivative of $5$ with respect to $x$ is $0$ .

$\frac{1}{x + 5} (1 + 0)$

Simplify the expression.

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Add $1$ and $0$ .

$\frac{1}{x + 5} \cdot 1$

Multiply $\frac{1}{x + 5}$ by $1$ .

$\frac{1}{x + 5}$

Find the Derivative f(x) = natural log of x+5

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