Calculus Examples

$f (x) = \frac{1}{2} e^{- x}$

Step 1

Find the first derivative.

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Step 1.1

Since $\frac{1}{2}$ is constant with respect to $x$ , the derivative of $\frac{1}{2} \cdot e^{- x}$ with respect to $x$ is $\frac{1}{2} \frac{d}{d x} [e^{- x}]$ .

$\frac{1}{2} \frac{d}{d x} [e^{- x}]$

Step 1.2

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 1.2.1

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

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

Step 1.2.2

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

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

Step 1.2.3

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

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

Step 1.3

Differentiate.

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Step 1.3.1

Combine $e^{- x}$ and $\frac{1}{2}$ .

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

Step 1.3.2

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

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

Step 1.3.3

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

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

Step 1.3.4

Combine fractions.

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Step 1.3.4.1

Multiply $- 1$ by $1$ .

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

Step 1.3.4.2

Combine $\frac{e^{- x}}{2}$ and $- 1$ .

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

Step 1.3.4.3

Simplify the expression.

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Step 1.3.4.3.1

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

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

Step 1.3.4.3.2

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

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

Step 1.3.4.3.3

Move the negative in front of the fraction.

$f' (x) = - \frac{e^{- x}}{2}$

Step 2

Find the second derivative.

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Step 2.1

Since $- \frac{1}{2}$ is constant with respect to $x$ , the derivative of $- \frac{e^{- x}}{2}$ with respect to $x$ is $- \frac{1}{2} \frac{d}{d x} [e^{- x}]$ .

$- \frac{1}{2} \frac{d}{d x} [e^{- x}]$

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 $g (x) = - x$ .

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

Differentiate.

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Step 2.3.1

Combine $e^{- x}$ and $\frac{1}{2}$ .

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

Step 2.3.2

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

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

Step 2.3.3

Multiply.

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Step 2.3.3.1

Multiply $- 1$ by $- 1$ .

$1 \frac{e^{- x}}{2} \frac{d}{d x} [x]$

Step 2.3.3.2

Multiply $\frac{e^{- x}}{2}$ by $1$ .

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $\frac{e^{- x}}{2}$ by $1$ .

$f'' (x) = \frac{e^{- x}}{2}$

Step 3

The second derivative of $f (x)$ with respect to $x$ is $\frac{e^{- x}}{2}$ .

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