Calculus Examples

$2^{2 x + 3}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = 2^{x}$ and $g (x) = 2 x + 3$ .

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Step 1.1

To apply the Chain Rule, set $u$ as $2 x + 3$ .

$\frac{d}{d u} [2^{u}] \frac{d}{d x} [2 x + 3]$

Step 1.2

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

$2^{u} \ln (2) \frac{d}{d x} [2 x + 3]$

Step 1.3

Replace all occurrences of $u$ with $2 x + 3$ .

$2^{2 x + 3} \ln (2) \frac{d}{d x} [2 x + 3]$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $2 x + 3$ with respect to $x$ is $\frac{d}{d x} [2 x] + \frac{d}{d x} [3]$ .

$2^{2 x + 3} \ln (2) (\frac{d}{d x} [2 x] + \frac{d}{d x} [3])$

Step 2.2

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

$2^{2 x + 3} \ln (2) (2 \frac{d}{d x} [x] + \frac{d}{d x} [3])$

Step 2.3

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

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

Step 2.4

Multiply $2$ by $1$ .

$2^{2 x + 3} \ln (2) (2 + \frac{d}{d x} [3])$

Step 2.5

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$2^{2 x + 3} \ln (2) (2 + 0)$

Step 2.6

Add $2$ and $0$ .

$2^{2 x + 3} \ln (2) \cdot 2$

Step 3

Raise $2$ to the power of $1$ .

$2^{1} \cdot 2^{2 x + 3} \ln (2)$

Step 4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$2^{1 + 2 x + 3} \ln (2)$

Step 5

Add $1$ and $3$ .

$2^{2 x + 4} \ln (2)$