Calculus Examples

$f (x) = \ln (e^{9 x} + 4)$

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 $g (x) = e^{9 x} + 4$ .

Tap for more steps...

Step 1.1

To apply the Chain Rule, set $u_{1}$ as $e^{9 x} + 4$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [e^{9 x} + 4]$

Step 1.2

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

$\frac{1}{u_{1}} \frac{d}{d x} [e^{9 x} + 4]$

Step 1.3

Replace all occurrences of $u_{1}$ with $e^{9 x} + 4$ .

$\frac{1}{e^{9 x} + 4} \frac{d}{d x} [e^{9 x} + 4]$

Step 2

By the Sum Rule, the derivative of $e^{9 x} + 4$ with respect to $x$ is $\frac{d}{d x} [e^{9 x}] + \frac{d}{d x} [4]$ .

$\frac{1}{e^{9 x} + 4} (\frac{d}{d x} [e^{9 x}] + \frac{d}{d x} [4])$

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 $g (x) = 9 x$ .

Tap for more steps...

Step 3.1

To apply the Chain Rule, set $u_{2}$ as $9 x$ .

$\frac{1}{e^{9 x} + 4} (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [9 x] + \frac{d}{d x} [4])$

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 $a$ = $e$ .

$\frac{1}{e^{9 x} + 4} (e^{u_{2}} \frac{d}{d x} [9 x] + \frac{d}{d x} [4])$

Step 3.3

Replace all occurrences of $u_{2}$ with $9 x$ .

$\frac{1}{e^{9 x} + 4} (e^{9 x} \frac{d}{d x} [9 x] + \frac{d}{d x} [4])$

Step 4

Differentiate.

Tap for more steps...

Step 4.1

Since $9$ is constant with respect to $x$ , the derivative of $9 x$ with respect to $x$ is $9 \frac{d}{d x} [x]$ .

$\frac{1}{e^{9 x} + 4} (e^{9 x} (9 \frac{d}{d x} [x]) + \frac{d}{d x} [4])$

Step 4.2

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

$\frac{1}{e^{9 x} + 4} (e^{9 x} (9 \cdot 1) + \frac{d}{d x} [4])$

Step 4.3

Simplify the expression.

Tap for more steps...

Step 4.3.1

Multiply $9$ by $1$ .

$\frac{1}{e^{9 x} + 4} (e^{9 x} \cdot 9 + \frac{d}{d x} [4])$

Step 4.3.2

Move $9$ to the left of $e^{9 x}$ .

$\frac{1}{e^{9 x} + 4} (9 \cdot e^{9 x} + \frac{d}{d x} [4])$

Step 4.4

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{1}{e^{9 x} + 4} (9 e^{9 x} + 0)$

Step 4.5

Combine fractions.

Tap for more steps...

Step 4.5.1

Add $9 e^{9 x}$ and $0$ .

$\frac{1}{e^{9 x} + 4} (9 e^{9 x})$

Step 4.5.2

Combine $9$ and $\frac{1}{e^{9 x} + 4}$ .

$\frac{9}{e^{9 x} + 4} e^{9 x}$

Step 4.5.3

Combine $\frac{9}{e^{9 x} + 4}$ and $e^{9 x}$ .

$\frac{9 e^{9 x}}{e^{9 x} + 4}$