Calculus Examples

${(\frac{b}{a + z^{2}})}^{9}$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d a} [f (g (a))]$ is $f' (g (a)) g' (a)$ where $f (a) = a^{9}$ and $g (a) = \frac{b}{a + z^{2}}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as $\frac{b}{a + z^{2}}$ .

$\frac{d}{d u_{1}} [{u_{1}}^{9}] \frac{d}{d a} [\frac{b}{a + z^{2}}]$

Step 1.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 9$ .

$9 {u_{1}}^{8} \frac{d}{d a} [\frac{b}{a + z^{2}}]$

Step 1.3

Replace all occurrences of $u_{1}$ with $\frac{b}{a + z^{2}}$ .

$9 {(\frac{b}{a + z^{2}})}^{8} \frac{d}{d a} [\frac{b}{a + z^{2}}]$

Step 2

Differentiate using the Constant Multiple Rule.

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Step 2.1

Since $b$ is constant with respect to $a$ , the derivative of $\frac{b}{a + z^{2}}$ with respect to $a$ is $b \frac{d}{d a} [\frac{1}{a + z^{2}}]$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b \frac{d}{d a} [\frac{1}{a + z^{2}}])$

Step 2.2

Rewrite $\frac{1}{a + z^{2}}$ as ${(a + z^{2})}^{- 1}$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b \frac{d}{d a} [{(a + z^{2})}^{- 1}])$

Step 3

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

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Step 3.1

To apply the Chain Rule, set $u_{2}$ as $a + z^{2}$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d a} [a + z^{2}]))$

Step 3.2

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

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {u_{2}}^{- 2} \frac{d}{d a} [a + z^{2}]))$

Step 3.3

Replace all occurrences of $u_{2}$ with $a + z^{2}$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {(a + z^{2})}^{- 2} \frac{d}{d a} [a + z^{2}]))$

Step 4

Differentiate.

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Step 4.1

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

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {(a + z^{2})}^{- 2} (\frac{d}{d a} [a] + \frac{d}{d a} [z^{2}])))$

Step 4.2

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

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {(a + z^{2})}^{- 2} (1 + \frac{d}{d a} [z^{2}])))$

Step 4.3

Since $z^{2}$ is constant with respect to $a$ , the derivative of $z^{2}$ with respect to $a$ is $0$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {(a + z^{2})}^{- 2} (1 + 0)))$

Step 4.4

Simplify the expression.

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Step 4.4.1

Add $1$ and $0$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {(a + z^{2})}^{- 2} \cdot 1))$

Step 4.4.2

Multiply $- 1$ by $1$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- {(a + z^{2})}^{- 2}))$

Step 5

Simplify.

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Step 5.1

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

$9 {(\frac{b}{a + z^{2}})}^{8} (b (- \frac{1}{{(a + z^{2})}^{2}}))$

Step 5.2

Combine $b$ and $\frac{1}{{(a + z^{2})}^{2}}$ .

$9 {(\frac{b}{a + z^{2}})}^{8} (- \frac{b}{{(a + z^{2})}^{2}})$

Step 6

Simplify.

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Step 6.1

Apply the product rule to $\frac{b}{a + z^{2}}$ .

$9 \frac{b^{8}}{{(a + z^{2})}^{8}} (- \frac{b}{{(a + z^{2})}^{2}})$

Step 6.2

Combine terms.

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Step 6.2.1

Combine $9$ and $\frac{b^{8}}{{(a + z^{2})}^{8}}$ .

$\frac{9 b^{8}}{{(a + z^{2})}^{8}} (- \frac{b}{{(a + z^{2})}^{2}})$

Step 6.2.2

Multiply $\frac{9 b^{8}}{{(a + z^{2})}^{8}}$ by $\frac{b}{{(a + z^{2})}^{2}}$ .

$- \frac{9 b^{8} b}{{(a + z^{2})}^{8} {(a + z^{2})}^{2}}$

Step 6.2.3

Raise $b$ to the power of $1$ .

$- \frac{9 (b^{1} b^{8})}{{(a + z^{2})}^{8} {(a + z^{2})}^{2}}$

Step 6.2.4

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

$- \frac{9 b^{1 + 8}}{{(a + z^{2})}^{8} {(a + z^{2})}^{2}}$

Step 6.2.5

Add $1$ and $8$ .

$- \frac{9 b^{9}}{{(a + z^{2})}^{8} {(a + z^{2})}^{2}}$

Step 6.2.6

Multiply ${(a + z^{2})}^{8}$ by ${(a + z^{2})}^{2}$ by adding the exponents.

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Step 6.2.6.1

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

$- \frac{9 b^{9}}{{(a + z^{2})}^{8 + 2}}$

Step 6.2.6.2

Add $8$ and $2$ .

$- \frac{9 b^{9}}{{(a + z^{2})}^{10}}$