Calculus Examples

$f (x) = \ln (e^{x} x^{3} {(x + 1)}^{4})$

Step 1

Let $x = f (x)$ , take the natural logarithm of both sides $\ln (x) = \ln (f (x))$ .

$\ln (f (x)) = \ln (\ln (e^{x} x^{3} {(x + 1)}^{4}))$

Step 2

Expand the right hand side.

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Step 2.1

Rewrite $\ln (e^{x} x^{3} {(x + 1)}^{4})$ as $\ln (e^{x} x^{3}) + \ln ({(x + 1)}^{4})$ .

$\ln (f (x)) = \ln (\ln (e^{x} x^{3}) + \ln ({(x + 1)}^{4}))$

Step 2.2

Rewrite $\ln (e^{x} x^{3})$ as $\ln (e^{x}) + \ln (x^{3})$ .

$\ln (f (x)) = \ln (\ln (e^{x}) + \ln (x^{3}) + \ln ({(x + 1)}^{4}))$

Step 2.3

Expand $\ln (e^{x})$ by moving $x$ outside the logarithm.

$\ln (f (x)) = \ln (x \ln (e) + \ln (x^{3}) + \ln ({(x + 1)}^{4}))$

Step 2.4

Expand $\ln (x^{3})$ by moving $3$ outside the logarithm.

$\ln (f (x)) = \ln (x \ln (e) + 3 \ln (x) + \ln ({(x + 1)}^{4}))$

Step 2.5

Expand $\ln ({(x + 1)}^{4})$ by moving $4$ outside the logarithm.

$\ln (f (x)) = \ln (x \ln (e) + 3 \ln (x) + 4 \ln (x + 1))$

Step 2.6

The natural logarithm of $e$ is $1$ .

$\ln (f (x)) = \ln (x \cdot 1 + 3 \ln (x) + 4 \ln (x + 1))$

Step 2.7

Multiply $x$ by $1$ .

$\ln (f (x)) = \ln (x + 3 \ln (x) + 4 \ln (x + 1))$

Step 3

Differentiate the expression using the chain rule, keeping in mind that $y$ is a function of $x$ .

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Step 3.1

Differentiate the left hand side $\ln (f (x))$ using the chain rule.

$\frac{x'}{x} = \ln (x + 3 \ln (x) + 4 \ln (x + 1))$

Step 3.2

Differentiate the right hand side.

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Step 3.2.1

Differentiate $\ln (x + 3 \ln (x) + 4 \ln (x + 1))$ .

$\frac{x'}{x} = \frac{d}{d x} [\ln (x + 3 \ln (x) + 4 \ln (x + 1))]$

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = \ln (x)$ and $g (x) = x + 3 \ln (x) + 4 \ln (x + 1)$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $x + 3 \ln (x) + 4 \ln (x + 1)$ .

$\frac{x'}{x} = \frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [x + 3 \ln (x) + 4 \ln (x + 1)]$

Step 3.2.2.2

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

$\frac{x'}{x} = \frac{1}{u_{1}} \frac{d}{d x} [x + 3 \ln (x) + 4 \ln (x + 1)]$

Step 3.2.2.3

Replace all occurrences of $u_{1}$ with $x + 3 \ln (x) + 4 \ln (x + 1)$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} \frac{d}{d x} [x + 3 \ln (x) + 4 \ln (x + 1)]$

Step 3.2.3

Differentiate.

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Step 3.2.3.1

By the Sum Rule, the derivative of $x + 3 \ln (x) + 4 \ln (x + 1)$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3 \ln (x)] + \frac{d}{d x} [4 \ln (x + 1)]$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{d}{d x} [x] + \frac{d}{d x} [3 \ln (x)] + \frac{d}{d x} [4 \ln (x + 1)])$

Step 3.2.3.2

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

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{d}{d x} [3 \ln (x)] + \frac{d}{d x} [4 \ln (x + 1)])$

Step 3.2.3.3

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

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + 3 \frac{d}{d x} [\ln (x)] + \frac{d}{d x} [4 \ln (x + 1)])$

Step 3.2.4

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

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + 3 \frac{1}{x} + \frac{d}{d x} [4 \ln (x + 1)])$

Step 3.2.5

Differentiate using the Constant Multiple Rule.

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Step 3.2.5.1

Combine $3$ and $\frac{1}{x}$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{d}{d x} [4 \ln (x + 1)])$

Step 3.2.5.2

Since $4$ is constant with respect to $x$ , the derivative of $4 \ln (x + 1)$ with respect to $x$ is $4 \frac{d}{d x} [\ln (x + 1)]$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + 4 \frac{d}{d x} [\ln (x + 1)])$

Step 3.2.6

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 $g (x) = x + 1$ .

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Step 3.2.6.1

To apply the Chain Rule, set $u_{2}$ as $x + 1$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + 4 (\frac{d}{d u_{2}} [\ln (u_{2})] \frac{d}{d x} [x + 1]))$

Step 3.2.6.2

The derivative of $\ln (u_{2})$ with respect to $u_{2}$ is $\frac{1}{u_{2}}$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + 4 (\frac{1}{u_{2}} \frac{d}{d x} [x + 1]))$

Step 3.2.6.3

Replace all occurrences of $u_{2}$ with $x + 1$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + 4 (\frac{1}{x + 1} \frac{d}{d x} [x + 1]))$

Step 3.2.7

Differentiate.

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Step 3.2.7.1

Combine $\frac{1}{x + 1}$ and $4$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{4}{x + 1} \frac{d}{d x} [x + 1])$

Step 3.2.7.2

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

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{4}{x + 1} (\frac{d}{d x} [x] + \frac{d}{d x} [1]))$

Step 3.2.7.3

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

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{4}{x + 1} (1 + \frac{d}{d x} [1]))$

Step 3.2.7.4

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

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{4}{x + 1} (1 + 0))$

Step 3.2.7.5

Combine into one fraction.

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Step 3.2.7.5.1

Add $1$ and $0$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{4}{x + 1} \cdot 1)$

Step 3.2.7.5.2

Simplify.

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Step 3.2.7.5.2.1

Multiply $\frac{4}{x + 1}$ by $1$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (1 + \frac{3}{x} + \frac{4}{x + 1})$

Step 3.2.7.5.2.2

Write $1$ as a fraction with a common denominator.

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3}{x} + \frac{x + 1}{x + 1} + \frac{4}{x + 1})$

Step 3.2.7.5.3

Combine the numerators over the common denominator.

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3}{x} + \frac{x + 1 + 4}{x + 1})$

Step 3.2.7.5.4

Add $1$ and $4$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3}{x} + \frac{x + 5}{x + 1})$

Step 3.2.8

To write $\frac{3}{x}$ as a fraction with a common denominator, multiply by $\frac{x + 1}{x + 1}$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3}{x} \cdot \frac{x + 1}{x + 1} + \frac{x + 5}{x + 1})$

Step 3.2.9

To write $\frac{x + 5}{x + 1}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3}{x} \cdot \frac{x + 1}{x + 1} + \frac{x + 5}{x + 1} \cdot \frac{x}{x})$

Step 3.2.10

Write each expression with a common denominator of $x (x + 1)$ , by multiplying each by an appropriate factor of $1$ .

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Step 3.2.10.1

Multiply $\frac{3}{x}$ by $\frac{x + 1}{x + 1}$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3 (x + 1)}{x (x + 1)} + \frac{x + 5}{x + 1} \cdot \frac{x}{x})$

Step 3.2.10.2

Multiply $\frac{x + 5}{x + 1}$ by $\frac{x}{x}$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3 (x + 1)}{x (x + 1)} + \frac{(x + 5) x}{(x + 1) x})$

Step 3.2.10.3

Reorder the factors of $(x + 1) x$ .

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} (\frac{3 (x + 1)}{x (x + 1)} + \frac{(x + 5) x}{x (x + 1)})$

Step 3.2.11

Combine the numerators over the common denominator.

$\frac{x'}{x} = \frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)} \cdot \frac{3 (x + 1) + (x + 5) x}{x (x + 1)}$

Step 3.2.12

Multiply $\frac{1}{x + 3 \ln (x) + 4 \ln (x + 1)}$ by $\frac{3 (x + 1) + (x + 5) x}{x (x + 1)}$ .

$\frac{x'}{x} = \frac{3 (x + 1) + (x + 5) x}{(x + 3 \ln (x) + 4 \ln (x + 1)) (x (x + 1))}$

Step 3.2.13

Simplify.

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Step 3.2.13.1

Apply the distributive property.

$\frac{x'}{x} = \frac{3 x + 3 \cdot 1 + (x + 5) x}{(x + 3 \ln (x) + 4 \ln (x + 1)) x (x + 1)}$

Step 3.2.13.2

Apply the distributive property.

$\frac{x'}{x} = \frac{3 x + 3 \cdot 1 + x \cdot x + 5 x}{(x + 3 \ln (x) + 4 \ln (x + 1)) x (x + 1)}$

Step 3.2.13.3

Apply the distributive property.

$\frac{x'}{x} = \frac{3 x + 3 \cdot 1 + x \cdot x + 5 x}{(x \cdot x + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.4

Simplify the numerator.

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Step 3.2.13.4.1

Simplify each term.

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Step 3.2.13.4.1.1

Multiply $3$ by $1$ .

$\frac{x'}{x} = \frac{3 x + 3 + x \cdot x + 5 x}{(x \cdot x + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.4.1.2

Multiply $x$ by $x$ .

$\frac{x'}{x} = \frac{3 x + 3 + x^{2} + 5 x}{(x \cdot x + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.4.2

Add $3 x$ and $5 x$ .

$\frac{x'}{x} = \frac{8 x + 3 + x^{2}}{(x \cdot x + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.5

Combine terms.

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Step 3.2.13.5.1

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

$\frac{x'}{x} = \frac{8 x + 3 + x^{2}}{(x^{1} x + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.5.2

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

$\frac{x'}{x} = \frac{8 x + 3 + x^{2}}{(x^{1} x^{1} + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.5.3

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

$\frac{x'}{x} = \frac{8 x + 3 + x^{2}}{(x^{1 + 1} + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.5.4

Add $1$ and $1$ .

$\frac{x'}{x} = \frac{8 x + 3 + x^{2}}{(x^{2} + 3 \ln (x) x + 4 \ln (x + 1) x) (x + 1)}$

Step 3.2.13.6

Reorder terms.

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) (x^{2} + 3 \ln (x) x + 4 \ln (x + 1) x)}$

Step 3.2.13.7

Factor $x$ out of $x^{2} + 3 \ln (x) x + 4 \ln (x + 1) x$ .

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Step 3.2.13.7.1

Factor $x$ out of $x^{2}$ .

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) (x \cdot x + 3 \ln (x) x + 4 \ln (x + 1) x)}$

Step 3.2.13.7.2

Factor $x$ out of $3 \ln (x) x$ .

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) (x \cdot x + x (3 \ln (x)) + 4 \ln (x + 1) x)}$

Step 3.2.13.7.3

Factor $x$ out of $4 \ln (x + 1) x$ .

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) (x \cdot x + x (3 \ln (x)) + x (4 \ln (x + 1)))}$

Step 3.2.13.7.4

Factor $x$ out of $x \cdot x + x (3 \ln (x))$ .

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) (x (x + 3 \ln (x)) + x (4 \ln (x + 1)))}$

Step 3.2.13.7.5

Factor $x$ out of $x (x + 3 \ln (x)) + x (4 \ln (x + 1))$ .

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) (x (x + 3 \ln (x) + 4 \ln (x + 1)))}$

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{(x + 1) x (x + 3 \ln (x) + 4 \ln (x + 1))}$

Step 3.2.13.8

Reorder factors in $\frac{x^{2} + 8 x + 3}{(x + 1) x (x + 3 \ln (x) + 4 \ln (x + 1))}$ .

$\frac{x'}{x} = \frac{x^{2} + 8 x + 3}{x (x + 1) (x + 3 \ln (x) + 4 \ln (x + 1))}$

Step 4

Isolate $x'$ and substitute the original function for $x$ in the right hand side.

$x' = \frac{x^{2} + 8 x + 3}{x (x + 1) (x + 3 \ln (x) + 4 \ln (x + 1))} \ln (e^{x} x^{3} {(x + 1)}^{4})$

Step 5

Simplify the right hand side.

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Step 5.1

Simplify the denominator.

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Step 5.1.1

Combine exponents.

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Step 5.1.1.1

Simplify $3 \ln (x)$ by moving $3$ inside the logarithm.

$x' = \frac{x^{2} + 8 x + 3}{x (x + 1) (x + \ln (x^{3}) + 4 \ln (x + 1))} \ln (e^{x} x^{3} {(x + 1)}^{4})$

Step 5.1.1.2

Simplify $4 \ln (x + 1)$ by moving $4$ inside the logarithm.

$x' = \frac{x^{2} + 8 x + 3}{x (x + 1) (x + \ln (x^{3}) + \ln ({(x + 1)}^{4}))} \ln (e^{x} x^{3} {(x + 1)}^{4})$

Step 5.1.2

Use the product property of logarithms, $\log_{b} (x) + \log_{b} (y) = \log_{b} (x y)$ .

$x' = \frac{x^{2} + 8 x + 3}{x (x + 1) (x + \ln (x^{3} {(x + 1)}^{4}))} \ln (e^{x} x^{3} {(x + 1)}^{4})$

Step 5.2

Combine $\frac{x^{2} + 8 x + 3}{x (x + 1) (x + \ln (x^{3} {(x + 1)}^{4}))}$ and $\ln (e^{x} x^{3} {(x + 1)}^{4})$ .

$x' = \frac{(x^{2} + 8 x + 3) \ln (e^{x} x^{3} {(x + 1)}^{4})}{x (x + 1) (x + \ln (x^{3} {(x + 1)}^{4}))}$

Step 5.3

Reorder factors in $\frac{(x^{2} + 8 x + 3) \ln (e^{x} x^{3} {(x + 1)}^{4})}{x (x + 1) (x + \ln (x^{3} {(x + 1)}^{4}))}$ .

$x' = \frac{\ln (x^{3} e^{x} {(x + 1)}^{4}) (x^{2} + 8 x + 3)}{x (x + 1) (x + \ln (x^{3} {(x + 1)}^{4}))}$