Calculus Examples

$e^{2 x} - x$

Step 1

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

$\frac{d}{d x} [e^{2 x}] + \frac{d}{d x} [- x]$

Step 2

Evaluate $\frac{d}{d x} [e^{2 x}]$ .

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

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

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

Step 2.1.3

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

$e^{2 x} \frac{d}{d x} [2 x] + \frac{d}{d x} [- x]$

Step 2.2

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

$e^{2 x} (2 \frac{d}{d x} [x]) + \frac{d}{d x} [- x]$

Step 2.3

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

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

Step 2.4

Multiply $2$ by $1$ .

$e^{2 x} \cdot 2 + \frac{d}{d x} [- x]$

Step 2.5

Move $2$ to the left of $e^{2 x}$ .

$2 e^{2 x} + \frac{d}{d x} [- x]$

Step 3

Evaluate $\frac{d}{d x} [- x]$ .

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Step 3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$2 e^{2 x} - \frac{d}{d x} [x]$

Step 3.2

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

$2 e^{2 x} - 1 \cdot 1$

Step 3.3

Multiply $- 1$ by $1$ .

$2 e^{2 x} - 1$