Calculus Examples

$\ln (e^{3 x} + x)$

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 $g (x) = e^{3 x} + x$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $e^{3 x} + x$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [e^{3 x} + x]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [e^{3 x} + x]$

Step 1.3

Replace all occurrences of $u_{1}$ with $e^{3 x} + x$ .

$\frac{1}{e^{3 x} + x} \frac{d}{d x} [e^{3 x} + x]$

Step 2

By the Sum Rule, the derivative of $e^{3 x} + x$ with respect to $x$ is $\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [x]$ .

$\frac{1}{e^{3 x} + x} (\frac{d}{d x} [e^{3 x}] + \frac{d}{d x} [x])$

Step 3

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = e^{x}$ and $g (x) = 3 x$ .

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $3 x$ .

$\frac{1}{e^{3 x} + x} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [3 x] + \frac{d}{d x} [x])$

Step 3.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$\frac{1}{e^{3 x} + x} (e^{u_{2}} \frac{d}{d x} [3 x] + \frac{d}{d x} [x])$

Step 3.3

Replace all occurrences of $u_{2}$ with $3 x$ .

$\frac{1}{e^{3 x} + x} (e^{3 x} \frac{d}{d x} [3 x] + \frac{d}{d x} [x])$

Step 4

Differentiate.

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Step 4.1

Since $3$ is constant with respect to $x$ , the derivative of $3 x$ with respect to $x$ is $3 \frac{d}{d x} [x]$ .

$\frac{1}{e^{3 x} + x} (e^{3 x} (3 \frac{d}{d x} [x]) + \frac{d}{d x} [x])$

Step 4.2

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

$\frac{1}{e^{3 x} + x} (e^{3 x} (3 \cdot 1) + \frac{d}{d x} [x])$

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $3$ by $1$ .

$\frac{1}{e^{3 x} + x} (e^{3 x} \cdot 3 + \frac{d}{d x} [x])$

Step 4.3.2

Move $3$ to the left of $e^{3 x}$ .

$\frac{1}{e^{3 x} + x} (3 \cdot e^{3 x} + \frac{d}{d x} [x])$

Step 4.4

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

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

Step 5

Simplify.

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Step 5.1

Reorder the factors of $\frac{1}{e^{3 x} + x} (3 e^{3 x} + 1)$ .

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

Step 5.2

Multiply $3 e^{3 x} + 1$ by $\frac{1}{e^{3 x} + x}$ .

$\frac{3 e^{3 x} + 1}{e^{3 x} + x}$