Find the Derivative - d/d@VAR f(x) = natural log of 6x+5e^x

$f (x) = \ln (6 x) + 5 e^{x}$

Step 1

By the Sum Rule, the derivative of $\ln (6 x) + 5 e^{x}$ with respect to $x$ is $\frac{d}{d x} [\ln (6 x)] + \frac{d}{d x} [5 e^{x}]$ .

$\frac{d}{d x} [\ln (6 x)] + \frac{d}{d x} [5 e^{x}]$

Step 2

Evaluate $\frac{d}{d x} [\ln (6 x)]$ .

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 $g (x) = 6 x$ .

To apply the Chain Rule, set $u$ as $6 x$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [6 x] + \frac{d}{d x} [5 e^{x}]$

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

$\frac{1}{u} \frac{d}{d x} [6 x] + \frac{d}{d x} [5 e^{x}]$

Replace all occurrences of $u$ with $6 x$ .

$\frac{1}{6 x} \frac{d}{d x} [6 x] + \frac{d}{d x} [5 e^{x}]$

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$\frac{1}{6 x} (6 \frac{d}{d x} [x]) + \frac{d}{d x} [5 e^{x}]$

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

$\frac{1}{6 x} (6 \cdot 1) + \frac{d}{d x} [5 e^{x}]$

Multiply $6$ by $1$ .

$\frac{1}{6 x} \cdot 6 + \frac{d}{d x} [5 e^{x}]$

Combine $\frac{1}{6 x}$ and $6$ .

$\frac{6}{6 x} + \frac{d}{d x} [5 e^{x}]$

Cancel the common factor of $6$ .

Cancel the common factor.

$\frac{6}{6 x} + \frac{d}{d x} [5 e^{x}]$

Rewrite the expression.

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

Step 3

Evaluate $\frac{d}{d x} [5 e^{x}]$ .

Since $5$ is constant with respect to $x$ , the derivative of $5 e^{x}$ with respect to $x$ is $5 \frac{d}{d x} [e^{x}]$ .

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

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

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