Calculus Examples

$\frac{1}{z - 1} - \frac{1}{z}$

Step 1

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

$\frac{d}{d z} [\frac{1}{z - 1}] + \frac{d}{d z} [- \frac{1}{z}]$

Step 2

Evaluate $\frac{d}{d z} [\frac{1}{z - 1}]$ .

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Step 2.1

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

$\frac{d}{d z} [{(z - 1)}^{- 1}] + \frac{d}{d z} [- \frac{1}{z}]$

Step 2.2

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

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Step 2.2.1

To apply the Chain Rule, set $u$ as $z - 1$ .

$\frac{d}{d u} [u^{- 1}] \frac{d}{d z} [z - 1] + \frac{d}{d z} [- \frac{1}{z}]$

Step 2.2.2

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

$- u^{- 2} \frac{d}{d z} [z - 1] + \frac{d}{d z} [- \frac{1}{z}]$

Step 2.2.3

Replace all occurrences of $u$ with $z - 1$ .

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

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

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

Step 2.6

Add $1$ and $0$ .

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

Step 2.7

Multiply $- 1$ by $1$ .

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

Step 3

Evaluate $\frac{d}{d z} [- \frac{1}{z}]$ .

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Step 3.1

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

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

Step 3.2

Rewrite $\frac{1}{z}$ as $z^{- 1}$ .

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

Step 3.3

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

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

Step 3.4

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

$- {(z - 1)}^{- 2} - - z^{- 2} + \frac{1}{z} \cdot 0$

Step 3.5

Multiply $- 1$ by $- 1$ .

$- {(z - 1)}^{- 2} + 1 z^{- 2} + \frac{1}{z} \cdot 0$

Step 3.6

Multiply $z^{- 2}$ by $1$ .

$- {(z - 1)}^{- 2} + z^{- 2} + \frac{1}{z} \cdot 0$

Step 3.7

Multiply $\frac{1}{z}$ by $0$ .

$- {(z - 1)}^{- 2} + z^{- 2} + 0$

Step 3.8

Add $z^{- 2}$ and $0$ .

$- {(z - 1)}^{- 2} + z^{- 2}$

Step 4

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

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

Step 5

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

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

Step 6

Reorder terms.

$\frac{1}{z^{2}} - \frac{1}{{(z - 1)}^{2}}$