Calculus Examples

$\sqrt{\frac{z - 5}{z + 5}}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{\frac{z - 5}{z + 5}}$ as ${(\frac{z - 5}{z + 5})}^{\frac{1}{2}}$ .

$\frac{d}{d z} [{(\frac{z - 5}{z + 5})}^{\frac{1}{2}}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d z} [f (g (z))]$ is $f' (g (z)) g' (z)$ where $f (z) = z^{\frac{1}{2}}$ and $g (z) = \frac{z - 5}{z + 5}$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $\frac{z - 5}{z + 5}$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 2.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 2.3

Replace all occurrences of $u$ with $\frac{z - 5}{z + 5}$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{\frac{1}{2} - 1} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 3

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

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 4

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

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 5

Combine the numerators over the common denominator.

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{\frac{1 - 2}{2}} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 6.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{\frac{- 1}{2}} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 7

Move the negative in front of the fraction.

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{d}{d z} [\frac{z - 5}{z + 5}]$

Step 8

Differentiate using the Quotient Rule which states that $\frac{d}{d z} [\frac{f (z)}{g (z)}]$ is $\frac{g (z) \frac{d}{d z} [f (z)] - f (z) \frac{d}{d z} [g (z)]}{{g (z)}^{2}}$ where $f (z) = z - 5$ and $g (z) = z + 5$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{(z + 5) \frac{d}{d z} [z - 5] - (z - 5) \frac{d}{d z} [z + 5]}{{(z + 5)}^{2}}$

Step 9

Differentiate.

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Step 9.1

By the Sum Rule, the derivative of $z - 5$ with respect to $z$ is $\frac{d}{d z} [z] + \frac{d}{d z} [- 5]$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{(z + 5) (\frac{d}{d z} [z] + \frac{d}{d z} [- 5]) - (z - 5) \frac{d}{d z} [z + 5]}{{(z + 5)}^{2}}$

Step 9.2

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

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{(z + 5) (1 + \frac{d}{d z} [- 5]) - (z - 5) \frac{d}{d z} [z + 5]}{{(z + 5)}^{2}}$

Step 9.3

Since $- 5$ is constant with respect to $z$ , the derivative of $- 5$ with respect to $z$ is $0$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{(z + 5) (1 + 0) - (z - 5) \frac{d}{d z} [z + 5]}{{(z + 5)}^{2}}$

Step 9.4

Simplify the expression.

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Step 9.4.1

Add $1$ and $0$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{(z + 5) \cdot 1 - (z - 5) \frac{d}{d z} [z + 5]}{{(z + 5)}^{2}}$

Step 9.4.2

Multiply $z + 5$ by $1$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{z + 5 - (z - 5) \frac{d}{d z} [z + 5]}{{(z + 5)}^{2}}$

Step 9.5

By the Sum Rule, the derivative of $z + 5$ with respect to $z$ is $\frac{d}{d z} [z] + \frac{d}{d z} [5]$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{z + 5 - (z - 5) (\frac{d}{d z} [z] + \frac{d}{d z} [5])}{{(z + 5)}^{2}}$

Step 9.6

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

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{z + 5 - (z - 5) (1 + \frac{d}{d z} [5])}{{(z + 5)}^{2}}$

Step 9.7

Since $5$ is constant with respect to $z$ , the derivative of $5$ with respect to $z$ is $0$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{z + 5 - (z - 5) (1 + 0)}{{(z + 5)}^{2}}$

Step 9.8

Combine fractions.

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Step 9.8.1

Add $1$ and $0$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{z + 5 - (z - 5) \cdot 1}{{(z + 5)}^{2}}$

Step 9.8.2

Multiply $- 1$ by $1$ .

$\frac{1}{2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}} \frac{z + 5 - (z - 5)}{{(z + 5)}^{2}}$

Step 9.8.3

Multiply $\frac{z + 5 - (z - 5)}{{(z + 5)}^{2}}$ by $\frac{1}{2}$ .

$\frac{z + 5 - (z - 5)}{{(z + 5)}^{2} \cdot 2} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}}$

Step 9.8.4

Move $2$ to the left of ${(z + 5)}^{2}$ .

$\frac{z + 5 - (z - 5)}{2 \cdot {(z + 5)}^{2}} {(\frac{z - 5}{z + 5})}^{- \frac{1}{2}}$

Step 10

Simplify.

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Step 10.1

Change the sign of the exponent by rewriting the base as its reciprocal.

$\frac{z + 5 - (z - 5)}{2 {(z + 5)}^{2}} {(\frac{z + 5}{z - 5})}^{\frac{1}{2}}$

Step 10.2

Apply the product rule to $\frac{z + 5}{z - 5}$ .

$\frac{z + 5 - (z - 5)}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.3

Apply the distributive property.

$\frac{z + 5 - z - - 5}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4

Combine terms.

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Step 10.4.1

Multiply $- 1$ by $- 5$ .

$\frac{z + 5 - z + 5}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.2

Subtract $z$ from $z$ .

$\frac{0 + 5 + 5}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.3

Add $0$ and $5$ .

$\frac{5 + 5}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.4

Add $5$ and $5$ .

$\frac{10}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.5

Cancel the common factor of $10$ and $2$ .

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Step 10.4.5.1

Factor $2$ out of $10$ .

$\frac{2 \cdot 5}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.5.2

Cancel the common factors.

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Step 10.4.5.2.1

Factor $2$ out of $2 {(z + 5)}^{2}$ .

$\frac{2 \cdot 5}{2 ({(z + 5)}^{2})} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.5.2.2

Cancel the common factor.

$\frac{2 \cdot 5}{2 {(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.5.2.3

Rewrite the expression.

$\frac{5}{{(z + 5)}^{2}} \cdot \frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$

Step 10.4.6

Multiply $\frac{5}{{(z + 5)}^{2}}$ by $\frac{{(z + 5)}^{\frac{1}{2}}}{{(z - 5)}^{\frac{1}{2}}}$ .

$\frac{5 {(z + 5)}^{\frac{1}{2}}}{{(z + 5)}^{2} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.7

Move ${(z + 5)}^{\frac{1}{2}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

$\frac{5}{{(z + 5)}^{2} {(z - 5)}^{\frac{1}{2}} {(z + 5)}^{- \frac{1}{2}}}$

Step 10.4.8

Multiply ${(z + 5)}^{2}$ by ${(z + 5)}^{- \frac{1}{2}}$ by adding the exponents.

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Step 10.4.8.1

Move ${(z + 5)}^{- \frac{1}{2}}$ .

$\frac{5}{{(z + 5)}^{- \frac{1}{2}} {(z + 5)}^{2} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.8.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{5}{{(z + 5)}^{- \frac{1}{2} + 2} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.8.3

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

$\frac{5}{{(z + 5)}^{- \frac{1}{2} + 2 \cdot \frac{2}{2}} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.8.4

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

$\frac{5}{{(z + 5)}^{- \frac{1}{2} + \frac{2 \cdot 2}{2}} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.8.5

Combine the numerators over the common denominator.

$\frac{5}{{(z + 5)}^{\frac{- 1 + 2 \cdot 2}{2}} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.8.6

Simplify the numerator.

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Step 10.4.8.6.1

Multiply $2$ by $2$ .

$\frac{5}{{(z + 5)}^{\frac{- 1 + 4}{2}} {(z - 5)}^{\frac{1}{2}}}$

Step 10.4.8.6.2

Add $- 1$ and $4$ .

$\frac{5}{{(z + 5)}^{\frac{3}{2}} {(z - 5)}^{\frac{1}{2}}}$