Calculus Examples

$y^{3} = e^{x} \ln (x^{2} - 1)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{3}) = \frac{d}{d x} (e^{x} \ln (x^{2} - 1))$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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Step 2.1.1

To apply the Chain Rule, set $u$ as $y$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d x} [y]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 3$ .

$3 u^{2} \frac{d}{d x} [y]$

Step 2.1.3

Replace all occurrences of $u$ with $y$ .

$3 y^{2} \frac{d}{d x} [y]$

Step 2.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$3 y^{2} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = e^{x}$ and $g (x) = \ln (x^{2} - 1)$ .

$e^{x} \frac{d}{d x} [\ln (x^{2} - 1)] + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.2

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 $g (x) = x^{2} - 1$ .

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Step 3.2.1

To apply the Chain Rule, set $u$ as $x^{2} - 1$ .

$e^{x} (\frac{d}{d u} [\ln (u)] \frac{d}{d x} [x^{2} - 1]) + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.2.2

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

$e^{x} (\frac{1}{u} \frac{d}{d x} [x^{2} - 1]) + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.2.3

Replace all occurrences of $u$ with $x^{2} - 1$ .

$e^{x} (\frac{1}{x^{2} - 1} \frac{d}{d x} [x^{2} - 1]) + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.3

Differentiate.

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Step 3.3.1

Combine $\frac{1}{x^{2} - 1}$ and $e^{x}$ .

$\frac{e^{x}}{x^{2} - 1} \frac{d}{d x} [x^{2} - 1] + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.3.2

By the Sum Rule, the derivative of $x^{2} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]$ .

$\frac{e^{x}}{x^{2} - 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [- 1]) + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.3.3

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

$\frac{e^{x}}{x^{2} - 1} (2 x + \frac{d}{d x} [- 1]) + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.3.4

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

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

Step 3.3.5

Combine fractions.

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Step 3.3.5.1

Add $2 x$ and $0$ .

$\frac{e^{x}}{x^{2} - 1} (2 x) + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.3.5.2

Combine $2$ and $\frac{e^{x}}{x^{2} - 1}$ .

$\frac{2 e^{x}}{x^{2} - 1} x + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.3.5.3

Combine $\frac{2 e^{x}}{x^{2} - 1}$ and $x$ .

$\frac{2 e^{x} x}{x^{2} - 1} + \ln (x^{2} - 1) \frac{d}{d x} [e^{x}]$

Step 3.4

Differentiate using the Exponential Rule which states that $\frac{d}{d x} [a^{x}]$ is $a^{x} \ln (a)$ where $a$ = $e$ .

$\frac{2 e^{x} x}{x^{2} - 1} + \ln (x^{2} - 1) e^{x}$

Step 3.5

To write $\ln (x^{2} - 1) e^{x}$ as a fraction with a common denominator, multiply by $\frac{x^{2} - 1}{x^{2} - 1}$ .

$\frac{2 e^{x} x}{x^{2} - 1} + \frac{\ln (x^{2} - 1) e^{x} (x^{2} - 1)}{x^{2} - 1}$

Step 3.6

Combine the numerators over the common denominator.

$\frac{2 e^{x} x + \ln (x^{2} - 1) e^{x} (x^{2} - 1)}{x^{2} - 1}$

Step 3.7

Simplify.

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Step 3.7.1

Simplify the numerator.

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Step 3.7.1.1

Simplify each term.

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Step 3.7.1.1.1

Apply the distributive property.

$\frac{2 e^{x} x + \ln (x^{2} - 1) e^{x} x^{2} + \ln (x^{2} - 1) e^{x} \cdot - 1}{x^{2} - 1}$

Step 3.7.1.1.2

Move $- 1$ to the left of $\ln (x^{2} - 1) e^{x}$ .

$\frac{2 e^{x} x + \ln (x^{2} - 1) e^{x} x^{2} - 1 \cdot (\ln (x^{2} - 1) e^{x})}{x^{2} - 1}$

Step 3.7.1.1.3

Rewrite $- 1 (\ln (x^{2} - 1) e^{x})$ as $- (\ln (x^{2} - 1) e^{x})$ .

$\frac{2 e^{x} x + \ln (x^{2} - 1) e^{x} x^{2} - \ln (x^{2} - 1) e^{x}}{x^{2} - 1}$

Step 3.7.1.2

Reorder factors in $2 e^{x} x + \ln (x^{2} - 1) e^{x} x^{2} - \ln (x^{2} - 1) e^{x}$ .

$\frac{2 x e^{x} + x^{2} e^{x} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{x^{2} - 1}$

Step 3.7.2

Reorder terms.

$\frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{x^{2} - 1}$

Step 4

Reform the equation by setting the left side equal to the right side.

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

Step 5

Divide each term in $3 y^{2} y' = \frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{x^{2} - 1}$ by $3 y^{2}$ and simplify.

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Step 5.1

Divide each term in $3 y^{2} y' = \frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{x^{2} - 1}$ by $3 y^{2}$ .

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

Step 5.2

Simplify the left side.

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Step 5.2.1

Cancel the common factor of $3$ .

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Step 5.2.1.1

Cancel the common factor.

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

Step 5.2.1.2

Rewrite the expression.

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

Step 5.2.2

Cancel the common factor of $y^{2}$ .

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Step 5.2.2.1

Cancel the common factor.

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

Step 5.2.2.2

Divide $y'$ by $1$ .

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Multiply the numerator by the reciprocal of the denominator.

$y' = \frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{x^{2} - 1} \cdot \frac{1}{3 y^{2}}$

Step 5.3.2

Simplify the denominator.

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Step 5.3.2.1

Rewrite $1$ as $1^{2}$ .

$y' = \frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{x^{2} - 1^{2}} \cdot \frac{1}{3 y^{2}}$

Step 5.3.2.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$y' = \frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{(x + 1) (x - 1)} \cdot \frac{1}{3 y^{2}}$

Step 5.3.3

Combine fractions.

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Step 5.3.3.1

Multiply $\frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{(x + 1) (x - 1)}$ by $\frac{1}{3 y^{2}}$ .

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

Step 5.3.3.2

Reorder.

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Step 5.3.3.2.1

Move $3$ to the left of $(x + 1) (x - 1)$ .

$y' = \frac{2 e^{x} x + e^{x} x^{2} \ln (x^{2} - 1) - e^{x} \ln (x^{2} - 1)}{3 \cdot ((x + 1) (x - 1)) y^{2}}$

Step 5.3.3.2.2

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

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

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = \frac{2 x e^{x} + x^{2} \ln (x^{2} - 1) e^{x} - \ln (x^{2} - 1) e^{x}}{3 y^{2} (x + 1) (x - 1)}$