Calculus Examples

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

Step 1

Rewrite $\frac{1}{1 - e^{- x}}$ as ${(1 - e^{- x})}^{- 1}$ .

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

Step 2

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

Tap for more steps...

Step 2.1

To apply the Chain Rule, set $u_{1}$ as $1 - e^{- x}$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $1 - e^{- x}$ .

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

Step 3

Differentiate.

Tap for more steps...

Step 3.1

By the Sum Rule, the derivative of $1 - e^{- x}$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- e^{- x}]$ .

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

Step 3.2

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

Multiply.

Tap for more steps...

Step 3.5.1

Multiply $- 1$ by $- 1$ .

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

Step 3.5.2

Multiply ${(1 - e^{- x})}^{- 2}$ by $1$ .

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

Step 4

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 $g (x) = - x$ .

Tap for more steps...

Step 4.1

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

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

Step 4.2

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

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

Step 4.3

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

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

Step 5

Differentiate.

Tap for more steps...

Step 5.1

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

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

Step 5.2

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

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

Step 5.3

Simplify the expression.

Tap for more steps...

Step 5.3.1

Multiply $- 1$ by $1$ .

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

Step 5.3.2

Move $- 1$ to the left of ${(1 - e^{- x})}^{- 2} e^{- x}$ .

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

Step 5.3.3

Rewrite $- 1 {(1 - e^{- x})}^{- 2}$ as $- {(1 - e^{- x})}^{- 2}$ .

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

Step 6

Simplify.

Tap for more steps...

Step 6.1

Reorder the factors of $- {(1 - e^{- x})}^{- 2} e^{- x}$ .

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

Step 6.2

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 6.3

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

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