Calculus Examples

$y = \sqrt{2 - e^{x}}$

Step 1

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

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

Step 2

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

Tap for more steps...

Step 2.1

To apply the Chain Rule, set $u$ as $2 - e^{x}$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $2 - e^{x}$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

Tap for more steps...

Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $1$ .

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

Step 7

Differentiate.

Tap for more steps...

Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

Combine fractions.

Tap for more steps...

Step 7.2.1

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

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

Step 7.2.2

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

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

Step 7.3

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

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

Step 7.4

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$\frac{1}{2 {(2 - e^{x})}^{\frac{1}{2}}} (0 + \frac{d}{d x} [- e^{x}])$

Step 7.5

Add $0$ and $\frac{d}{d x} [- e^{x}]$ .

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

Step 7.6

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

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

Step 8

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

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

Step 9

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

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