Calculus Examples

$\sqrt{x} e^{- x}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$ .

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

Step 2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = e^{- x}$ .

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

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 $g (x) = - x$ .

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 4

Differentiate.

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Step 4.1

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

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

Step 4.2

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

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

Step 4.3

Simplify the expression.

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Step 4.3.1

Multiply $- 1$ by $1$ .

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

Step 4.3.2

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

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

Step 4.3.3

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

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

Step 4.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = \frac{1}{2}$ .

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

Step 5

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 6

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 7

Combine the numerators over the common denominator.

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

Step 8

Simplify the numerator.

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Step 8.1

Multiply $- 1$ by $2$ .

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

Step 8.2

Subtract $2$ from $1$ .

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

Step 9

Move the negative in front of the fraction.

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

Step 10

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

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

Step 11

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

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

Step 12

Simplify the expression.

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Step 12.1

Move $x^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 12.2

Reorder terms.

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