Calculus Examples

$y = \ln ({(3 x^{4} + 5 x)}^{\frac{9}{7}})$

Step 1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = {(3 x^{4} + 5 x)}^{\frac{9}{7}}$ .

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Step 1.1

To apply the Chain Rule, set $u_{1}$ as ${(3 x^{4} + 5 x)}^{\frac{9}{7}}$ .

$\frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [{(3 x^{4} + 5 x)}^{\frac{9}{7}}]$

Step 1.2

The derivative of $\ln (u_{1})$ with respect to $u_{1}$ is $\frac{1}{u_{1}}$ .

$\frac{1}{u_{1}} \frac{d}{d x} [{(3 x^{4} + 5 x)}^{\frac{9}{7}}]$

Step 1.3

Replace all occurrences of $u_{1}$ with ${(3 x^{4} + 5 x)}^{\frac{9}{7}}$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} \frac{d}{d x} [{(3 x^{4} + 5 x)}^{\frac{9}{7}}]$

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{9}{7}}$ and $g (x) = 3 x^{4} + 5 x$ .

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Step 2.1

To apply the Chain Rule, set $u_{2}$ as $3 x^{4} + 5 x$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{d}{d u_{2}} [{u_{2}}^{\frac{9}{7}}] \frac{d}{d x} [3 x^{4} + 5 x])$

Step 2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = \frac{9}{7}$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {u_{2}}^{\frac{9}{7} - 1} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 2.3

Replace all occurrences of $u_{2}$ with $3 x^{4} + 5 x$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {(3 x^{4} + 5 x)}^{\frac{9}{7} - 1} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{7}{7}$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {(3 x^{4} + 5 x)}^{\frac{9}{7} - 1 \cdot \frac{7}{7}} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 4

Combine $- 1$ and $\frac{7}{7}$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {(3 x^{4} + 5 x)}^{\frac{9}{7} + \frac{- 1 \cdot 7}{7}} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 5

Combine the numerators over the common denominator.

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {(3 x^{4} + 5 x)}^{\frac{9 - 1 \cdot 7}{7}} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $7$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {(3 x^{4} + 5 x)}^{\frac{9 - 7}{7}} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 6.2

Subtract $7$ from $9$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9}{7} {(3 x^{4} + 5 x)}^{\frac{2}{7}} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 7

Combine $\frac{9}{7}$ and ${(3 x^{4} + 5 x)}^{\frac{2}{7}}$ .

$\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}} (\frac{9 {(3 x^{4} + 5 x)}^{\frac{2}{7}}}{7} \frac{d}{d x} [3 x^{4} + 5 x])$

Step 8

Multiply $\frac{9 {(3 x^{4} + 5 x)}^{\frac{2}{7}}}{7}$ by $\frac{1}{{(3 x^{4} + 5 x)}^{\frac{9}{7}}}$ .

$\frac{9 {(3 x^{4} + 5 x)}^{\frac{2}{7}}}{7 {(3 x^{4} + 5 x)}^{\frac{9}{7}}} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 9

Move ${(3 x^{4} + 5 x)}^{\frac{2}{7}}$ to the denominator using the negative exponent rule $b^{n} = \frac{1}{b^{- n}}$ .

$\frac{9}{7 {(3 x^{4} + 5 x)}^{\frac{9}{7}} {(3 x^{4} + 5 x)}^{- \frac{2}{7}}} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 10

Simplify the denominator.

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Step 10.1

Multiply ${(3 x^{4} + 5 x)}^{\frac{9}{7}}$ by ${(3 x^{4} + 5 x)}^{- \frac{2}{7}}$ by adding the exponents.

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Step 10.1.1

Move ${(3 x^{4} + 5 x)}^{- \frac{2}{7}}$ .

$\frac{9}{7 ({(3 x^{4} + 5 x)}^{- \frac{2}{7}} {(3 x^{4} + 5 x)}^{\frac{9}{7}})} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 10.1.2

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

$\frac{9}{7 {(3 x^{4} + 5 x)}^{- \frac{2}{7} + \frac{9}{7}}} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 10.1.3

Combine the numerators over the common denominator.

$\frac{9}{7 {(3 x^{4} + 5 x)}^{\frac{- 2 + 9}{7}}} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 10.1.4

Add $- 2$ and $9$ .

$\frac{9}{7 {(3 x^{4} + 5 x)}^{\frac{7}{7}}} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 10.1.5

Divide $7$ by $7$ .

$\frac{9}{7 {(3 x^{4} + 5 x)}^{1}} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 10.2

Simplify $7 {(3 x^{4} + 5 x)}^{1}$ .

$\frac{9}{7 (3 x^{4} + 5 x)} \frac{d}{d x} [3 x^{4} + 5 x]$

Step 11

By the Sum Rule, the derivative of $3 x^{4} + 5 x$ with respect to $x$ is $\frac{d}{d x} [3 x^{4}] + \frac{d}{d x} [5 x]$ .

$\frac{9}{7 (3 x^{4} + 5 x)} (\frac{d}{d x} [3 x^{4}] + \frac{d}{d x} [5 x])$

Step 12

Since $3$ is constant with respect to $x$ , the derivative of $3 x^{4}$ with respect to $x$ is $3 \frac{d}{d x} [x^{4}]$ .

$\frac{9}{7 (3 x^{4} + 5 x)} (3 \frac{d}{d x} [x^{4}] + \frac{d}{d x} [5 x])$

Step 13

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$\frac{9}{7 (3 x^{4} + 5 x)} (3 (4 x^{3}) + \frac{d}{d x} [5 x])$

Step 14

Multiply $4$ by $3$ .

$\frac{9}{7 (3 x^{4} + 5 x)} (12 x^{3} + \frac{d}{d x} [5 x])$

Step 15

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

$\frac{9}{7 (3 x^{4} + 5 x)} (12 x^{3} + 5 \frac{d}{d x} [x])$

Step 16

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

$\frac{9}{7 (3 x^{4} + 5 x)} (12 x^{3} + 5 \cdot 1)$

Step 17

Multiply $5$ by $1$ .

$\frac{9}{7 (3 x^{4} + 5 x)} (12 x^{3} + 5)$

Step 18

Simplify.

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Step 18.1

Apply the distributive property.

$\frac{9}{7 (3 x^{4}) + 7 (5 x)} (12 x^{3} + 5)$

Step 18.2

Combine terms.

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Step 18.2.1

Multiply $3$ by $7$ .

$\frac{9}{21 x^{4} + 7 (5 x)} (12 x^{3} + 5)$

Step 18.2.2

Multiply $5$ by $7$ .

$\frac{9}{21 x^{4} + 35 x} (12 x^{3} + 5)$

Step 18.3

Reorder the factors of $\frac{9}{21 x^{4} + 35 x} (12 x^{3} + 5)$ .

$(12 x^{3} + 5) \frac{9}{21 x^{4} + 35 x}$

Step 18.4

Factor $7 x$ out of $21 x^{4} + 35 x$ .

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Step 18.4.1

Factor $7 x$ out of $21 x^{4}$ .

$(12 x^{3} + 5) \frac{9}{7 x (3 x^{3}) + 35 x}$

Step 18.4.2

Factor $7 x$ out of $35 x$ .

$(12 x^{3} + 5) \frac{9}{7 x (3 x^{3}) + 7 x (5)}$

Step 18.4.3

Factor $7 x$ out of $7 x (3 x^{3}) + 7 x (5)$ .

$(12 x^{3} + 5) \frac{9}{7 x (3 x^{3} + 5)}$

Step 18.5

Multiply $12 x^{3} + 5$ by $\frac{9}{7 x (3 x^{3} + 5)}$ .

$\frac{(12 x^{3} + 5) \cdot 9}{7 x (3 x^{3} + 5)}$

Step 18.6

Move $9$ to the left of $12 x^{3} + 5$ .

$\frac{9 (12 x^{3} + 5)}{7 x (3 x^{3} + 5)}$