Calculus Examples

$2 e^{2 x} - e^{- x}$

Step 1

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

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

Step 2

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

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Step 2.1

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

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

Step 2.2

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 2.2.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

Step 2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

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

Step 2.2.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Multiply $2$ by $1$ .

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

Step 2.6

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

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

Step 2.7

Multiply $2$ by $2$ .

$4 e^{2 x} + \frac{d}{d x} [- e^{- x}]$

Step 3

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

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Step 3.1

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

$4 e^{2 x} - \frac{d}{d x} [e^{- x}]$

Step 3.2

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 $g (x) = - x$ .

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Step 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

$4 e^{2 x} - (e^{- x} \frac{d}{d x} [- x])$

Step 3.3

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

$4 e^{2 x} - (e^{- x} (- \frac{d}{d x} [x]))$

Step 3.4

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

$4 e^{2 x} - (e^{- x} (- 1 \cdot 1))$

Step 3.5

Multiply $- 1$ by $1$ .

$4 e^{2 x} - (e^{- x} \cdot - 1)$

Step 3.6

Move $- 1$ to the left of $e^{- x}$ .

$4 e^{2 x} - (- 1 \cdot e^{- x})$

Step 3.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$4 e^{2 x} - - e^{- x}$

Step 3.8

Multiply $- 1$ by $- 1$ .

$4 e^{2 x} + 1 e^{- x}$

Step 3.9

Multiply $e^{- x}$ by $1$ .

$4 e^{2 x} + e^{- x}$