Calculus Examples

$\frac{\sinh (x)}{e^{x}}$

Step 1

Consider the limit definition of the derivative.

$f' (x) = lim_{h \to 0} \frac{f (x + h) - f (x)}{h}$

Step 2

Find the components of the definition.

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Step 2.1

Evaluate the function at $x = x + h$ .

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Step 2.1.1

Replace the variable $x$ with $x + h$ in the expression.

$f (x + h) = \frac{\sinh (x + h)}{e^{x + h}}$

Step 2.1.2

The final answer is $\frac{\sinh (x + h)}{e^{x + h}}$ .

$\frac{\sinh (x + h)}{e^{x + h}}$

Step 2.2

Find the components of the definition.

$f (x + h) = \frac{\sinh (x + h)}{e^{x + h}}$

$f (x) = \frac{\sinh (x)}{e^{x}}$

$f (x + h) = \frac{\sinh (x + h)}{e^{x + h}}$

$f (x) = \frac{\sinh (x)}{e^{x}}$

Step 3

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{\frac{\sinh (x + h)}{e^{x + h}} - (\frac{\sinh (x)}{e^{x}})}{h}$

Step 4

Combine terms.

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Step 4.1

To write $\frac{\sinh (x + h)}{e^{x + h}}$ as a fraction with a common denominator, multiply by $\frac{e^{x}}{e^{x}}$ .

$lim_{h \to 0} \frac{\frac{\sinh (x + h)}{e^{x + h}} \cdot \frac{e^{x}}{e^{x}} - (\frac{\sinh (x)}{e^{x}})}{h}$

Step 4.2

To write $- (\frac{\sinh (x)}{e^{x}})$ as a fraction with a common denominator, multiply by $\frac{e^{h + x}}{e^{h + x}}$ .

$lim_{h \to 0} \frac{\frac{\sinh (x + h)}{e^{x + h}} \cdot \frac{e^{x}}{e^{x}} - (\frac{\sinh (x)}{e^{x}}) \cdot \frac{e^{h + x}}{e^{h + x}}}{h}$

Step 4.3

Write each expression with a common denominator of $e^{x + h} e^{x}$ , by multiplying each by an appropriate factor of $1$ .

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Step 4.3.1

Multiply $\frac{\sinh (x + h)}{e^{x + h}}$ by $\frac{e^{x}}{e^{x}}$ .

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x}}{e^{x + h} e^{x}} - (\frac{\sinh (x)}{e^{x}}) \cdot \frac{e^{h + x}}{e^{h + x}}}{h}$

Step 4.3.2

Multiply $e^{x + h}$ by $e^{x}$ by adding the exponents.

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Step 4.3.2.1

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

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x}}{e^{x + h + x}} - (\frac{\sinh (x)}{e^{x}}) \cdot \frac{e^{h + x}}{e^{h + x}}}{h}$

Step 4.3.2.2

Add $x$ and $x$ .

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x}}{e^{h + 2 x}} - (\frac{\sinh (x)}{e^{x}}) \cdot \frac{e^{h + x}}{e^{h + x}}}{h}$

Step 4.3.3

Multiply $\frac{\sinh (x)}{e^{x}}$ by $\frac{e^{h + x}}{e^{h + x}}$ .

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x}}{e^{h + 2 x}} - \frac{\sinh (x) e^{h + x}}{e^{x} e^{h + x}}}{h}$

Step 4.3.4

Multiply $e^{x}$ by $e^{h + x}$ by adding the exponents.

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Step 4.3.4.1

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

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x}}{e^{h + 2 x}} - \frac{\sinh (x) e^{h + x}}{e^{x + h + x}}}{h}$

Step 4.3.4.2

Add $x$ and $x$ .

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x}}{e^{h + 2 x}} - \frac{\sinh (x) e^{h + x}}{e^{h + 2 x}}}{h}$

Step 4.4

Combine the numerators over the common denominator.

$lim_{h \to 0} \frac{\frac{\sinh (x + h) e^{x} - \sinh (x) e^{h + x}}{e^{h + 2 x}}}{h}$

Step 5

Simplify the limit argument.

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Step 5.1

Multiply the numerator by the reciprocal of the denominator.

$lim_{h \to 0} \frac{\sinh (x + h) e^{x} - \sinh (x) e^{h + x}}{e^{h + 2 x}} \cdot \frac{1}{h}$

Step 5.2

Multiply $\frac{\sinh (x + h) e^{x} - \sinh (x) e^{h + x}}{e^{h + 2 x}}$ by $\frac{1}{h}$ .

$lim_{h \to 0} \frac{\sinh (x + h) e^{x} - \sinh (x) e^{h + x}}{e^{h + 2 x} h}$

Step 6

Apply L'Hospital's rule.

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Step 6.1

Evaluate the limit of the numerator and the limit of the denominator.

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Step 6.1.1

Take the limit of the numerator and the limit of the denominator.

$\frac{lim_{h \to 0} \sinh (x + h) e^{x} - \sinh (x) e^{h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2

Evaluate the limit of the numerator.

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Step 6.1.2.1

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{lim_{h \to 0} \sinh (x + h) e^{x} - lim_{h \to 0} \sinh (x) e^{h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.2

Move the term $e^{x}$ outside of the limit because it is constant with respect to $h$ .

$\frac{e^{x} lim_{h \to 0} \sinh (x + h) - lim_{h \to 0} \sinh (x) e^{h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.3

Move the term $\sinh (x)$ outside of the limit because it is constant with respect to $h$ .

$\frac{e^{x} lim_{h \to 0} \sinh (x + h) - \sinh (x) lim_{h \to 0} e^{h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.4

Move the limit into the exponent.

$\frac{e^{x} lim_{h \to 0} \sinh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.5

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \sinh (x + h) - \sinh (x) e^{lim_{h \to 0} h + lim_{h \to 0} x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.6

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \sinh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.7

Evaluate the limits by plugging in $0$ for all occurrences of $h$ .

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Step 6.1.2.7.1

Evaluate the limit of $\sinh (x + h)$ by plugging in $0$ for $h$ .

$\frac{e^{x} \sinh (x + 0) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.7.2

Add $x$ and $0$ .

$\frac{e^{x} \sinh (x) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.7.3

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{e^{x} \sinh (x) - \sinh (x) e^{0 + x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.8

Combine the opposite terms in $e^{x} \sinh (x) - \sinh (x) e^{0 + x}$ .

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Step 6.1.2.8.1

Add $0$ and $x$ .

$\frac{e^{x} \sinh (x) - \sinh (x) e^{x}}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.8.2

Reorder the factors in the terms $e^{x} \sinh (x)$ and $- \sinh (x) e^{x}$ .

$\frac{e^{x} \sinh (x) - e^{x} \sinh (x)}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.2.8.3

Subtract $e^{x} \sinh (x)$ from $e^{x} \sinh (x)$ .

$\frac{0}{lim_{h \to 0} e^{h + 2 x} h}$

Step 6.1.3

Evaluate the limit of the denominator.

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Step 6.1.3.1

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{0}{lim_{h \to 0} e^{h + 2 x} \cdot lim_{h \to 0} h}$

Step 6.1.3.2

Move the limit into the exponent.

$\frac{0}{e^{lim_{h \to 0} h + 2 x} \cdot lim_{h \to 0} h}$

Step 6.1.3.3

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{0}{e^{lim_{h \to 0} h + lim_{h \to 0} 2 x} \cdot lim_{h \to 0} h}$

Step 6.1.3.4

Evaluate the limit of $2 x$ which is constant as $h$ approaches $0$ .

$\frac{0}{e^{lim_{h \to 0} h + 2 x} \cdot lim_{h \to 0} h}$

Step 6.1.3.5

Evaluate the limits by plugging in $0$ for all occurrences of $h$ .

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Step 6.1.3.5.1

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{0}{e^{0 + 2 x} \cdot lim_{h \to 0} h}$

Step 6.1.3.5.2

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{0}{e^{0 + 2 x} \cdot 0}$

Step 6.1.3.6

Simplify the answer.

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Step 6.1.3.6.1

Add $0$ and $2 x$ .

$\frac{0}{e^{2 x} \cdot 0}$

Step 6.1.3.6.2

Multiply $e^{2 x}$ by $0$ .

$\frac{0}{0}$

Step 6.1.3.6.3

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 6.1.3.7

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 6.1.4

The expression contains a division by $0$ . The expression is undefined.

Undefined

$\frac{0}{0}$

Step 6.2

Since $\frac{0}{0}$ is of indeterminate form, apply L'Hospital's Rule. L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.

$lim_{h \to 0} \frac{\sinh (x + h) e^{x} - \sinh (x) e^{h + x}}{e^{h + 2 x} h} = lim_{h \to 0} \frac{\frac{d}{d h} [\sinh (x + h) e^{x} - \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3

Find the derivative of the numerator and denominator.

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Step 6.3.1

Differentiate the numerator and denominator.

$lim_{h \to 0} \frac{\frac{d}{d h} [\sinh (x + h) e^{x} - \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.2

By the Sum Rule, the derivative of $\sinh (x + h) e^{x} - \sinh (x) e^{h + x}$ with respect to $h$ is $\frac{d}{d h} [\sinh (x + h) e^{x}] + \frac{d}{d h} [- \sinh (x) e^{h + x}]$ .

$lim_{h \to 0} \frac{\frac{d}{d h} [\sinh (x + h) e^{x}] + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3

Evaluate $\frac{d}{d h} [\sinh (x + h) e^{x}]$ .

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Step 6.3.3.1

Since $e^{x}$ is constant with respect to $h$ , the derivative of $\sinh (x + h) e^{x}$ with respect to $h$ is $e^{x} \frac{d}{d h} [\sinh (x + h)]$ .

$lim_{h \to 0} \frac{e^{x} \frac{d}{d h} [\sinh (x + h)] + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.2

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

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Step 6.3.3.2.1

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

$lim_{h \to 0} \frac{e^{x} (\frac{d}{d u_{1}} [\sinh (u_{1})] \frac{d}{d h} [x + h]) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.2.2

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

$lim_{h \to 0} \frac{e^{x} (\cosh (u_{1}) \frac{d}{d h} [x + h]) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.2.3

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

$lim_{h \to 0} \frac{e^{x} (\cosh (x + h) \frac{d}{d h} [x + h]) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.3

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

$lim_{h \to 0} \frac{e^{x} (\cosh (x + h) (\frac{d}{d h} [x] + \frac{d}{d h} [h])) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.4

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

$lim_{h \to 0} \frac{e^{x} (\cosh (x + h) (0 + \frac{d}{d h} [h])) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.5

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

$lim_{h \to 0} \frac{e^{x} (\cosh (x + h) (0 + 1)) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.6

Add $0$ and $1$ .

$lim_{h \to 0} \frac{e^{x} (\cosh (x + h) \cdot 1) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.3.7

Multiply $\cosh (x + h)$ by $1$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) + \frac{d}{d h} [- \sinh (x) e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4

Evaluate $\frac{d}{d h} [- \sinh (x) e^{h + x}]$ .

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Step 6.3.4.1

Since $- \sinh (x)$ is constant with respect to $h$ , the derivative of $- \sinh (x) e^{h + x}$ with respect to $h$ is $- \sinh (x) \frac{d}{d h} [e^{h + x}]$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) \frac{d}{d h} [e^{h + x}]}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.2

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

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Step 6.3.4.2.1

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d h} [h + x])}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{2}} [a^{u_{2}}]$ is $a^{u_{2}} \ln (a)$ where $a$ = $e$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (e^{u_{2}} \frac{d}{d h} [h + x])}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.2.3

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (e^{h + x} \frac{d}{d h} [h + x])}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.3

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (e^{h + x} (\frac{d}{d h} [h] + \frac{d}{d h} [x]))}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.4

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (e^{h + x} (1 + \frac{d}{d h} [x]))}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.5

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (e^{h + x} (1 + 0))}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.6

Add $1$ and $0$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) (e^{h + x} \cdot 1)}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.4.7

Multiply $e^{h + x}$ by $1$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - \sinh (x) e^{h + x}}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.5

Reorder terms.

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{\frac{d}{d h} [e^{h + 2 x} h]}$

Step 6.3.6

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} \frac{d}{d h} [h] + h \frac{d}{d h} [e^{h + 2 x}]}$

Step 6.3.7

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} \cdot 1 + h \frac{d}{d h} [e^{h + 2 x}]}$

Step 6.3.8

Multiply $e^{h + 2 x}$ by $1$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h \frac{d}{d h} [e^{h + 2 x}]}$

Step 6.3.9

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

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Step 6.3.9.1

To apply the Chain Rule, set $u$ as $h + 2 x$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (\frac{d}{d u} [e^{u}] \frac{d}{d h} [h + 2 x])}$

Step 6.3.9.2

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (e^{u} \frac{d}{d h} [h + 2 x])}$

Step 6.3.9.3

Replace all occurrences of $u$ with $h + 2 x$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (e^{h + 2 x} \frac{d}{d h} [h + 2 x])}$

Step 6.3.10

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (e^{h + 2 x} (\frac{d}{d h} [h] + \frac{d}{d h} [2 x]))}$

Step 6.3.11

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

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (e^{h + 2 x} (1 + \frac{d}{d h} [2 x]))}$

Step 6.3.12

Since $2 x$ is constant with respect to $h$ , the derivative of $2 x$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (e^{h + 2 x} (1 + 0))}$

Step 6.3.13

Add $1$ and $0$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h (e^{h + 2 x} \cdot 1)}$

Step 6.3.14

Multiply $e^{h + 2 x}$ by $1$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} + h e^{h + 2 x}}$

Step 6.3.15

Simplify.

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Step 6.3.15.1

Reorder terms.

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{e^{h + 2 x} h + e^{h + 2 x}}$

Step 6.3.15.2

Reorder factors in $e^{h + 2 x} h + e^{h + 2 x}$ .

$lim_{h \to 0} \frac{e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{h e^{h + 2 x} + e^{h + 2 x}}$

Step 7

Evaluate the limit.

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Step 7.1

Split the limit using the Limits Quotient Rule on the limit as $h$ approaches $0$ .

$\frac{lim_{h \to 0} e^{x} \cosh (x + h) - e^{h + x} \sinh (x)}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.2

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{lim_{h \to 0} e^{x} \cosh (x + h) - lim_{h \to 0} e^{h + x} \sinh (x)}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.3

Move the term $e^{x}$ outside of the limit because it is constant with respect to $h$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - lim_{h \to 0} e^{h + x} \sinh (x)}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.4

Move the term $\sinh (x)$ outside of the limit because it is constant with respect to $h$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) lim_{h \to 0} e^{h + x}}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.5

Move the limit into the exponent.

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.6

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + lim_{h \to 0} x}}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.7

Evaluate the limit of $x$ which is constant as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h e^{h + 2 x} + e^{h + 2 x}}$

Step 7.8

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h e^{h + 2 x} + lim_{h \to 0} e^{h + 2 x}}$

Step 7.9

Split the limit using the Product of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot lim_{h \to 0} e^{h + 2 x} + lim_{h \to 0} e^{h + 2 x}}$

Step 7.10

Move the limit into the exponent.

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + lim_{h \to 0} e^{h + 2 x}}$

Step 7.11

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + lim_{h \to 0} 2 x} + lim_{h \to 0} e^{h + 2 x}}$

Step 7.12

Evaluate the limit of $2 x$ which is constant as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + lim_{h \to 0} e^{h + 2 x}}$

Step 7.13

Move the limit into the exponent.

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 7.14

Split the limit using the Sum of Limits Rule on the limit as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + lim_{h \to 0} 2 x}}$

Step 7.15

Evaluate the limit of $2 x$ which is constant as $h$ approaches $0$ .

$\frac{e^{x} lim_{h \to 0} \cosh (x + h) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 8

Evaluate the limits by plugging in $0$ for all occurrences of $h$ .

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Step 8.1

Evaluate the limit of $\cosh (x + h)$ by plugging in $0$ for $h$ .

$\frac{e^{x} \cosh (x + 0) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 8.2

Add $x$ and $0$ .

$\frac{e^{x} \cosh (x) - \sinh (x) e^{lim_{h \to 0} h + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 8.3

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{e^{x} \cosh (x) - \sinh (x) e^{0 + x}}{lim_{h \to 0} h \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 8.4

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{e^{x} \cosh (x) - \sinh (x) e^{0 + x}}{0 \cdot e^{lim_{h \to 0} h + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 8.5

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{e^{x} \cosh (x) - \sinh (x) e^{0 + x}}{0 \cdot e^{0 + 2 x} + e^{lim_{h \to 0} h + 2 x}}$

Step 8.6

Evaluate the limit of $h$ by plugging in $0$ for $h$ .

$\frac{e^{x} \cosh (x) - \sinh (x) e^{0 + x}}{0 \cdot e^{0 + 2 x} + e^{0 + 2 x}}$

Step 9

Simplify the answer.

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Step 9.1

Simplify the numerator.

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Step 9.1.1

Add $0$ and $x$ .

$\frac{e^{x} \cosh (x) - \sinh (x) e^{x}}{0 \cdot e^{0 + 2 x} + e^{0 + 2 x}}$

Step 9.1.2

Factor $e^{x}$ out of $e^{x} \cosh (x) - \sinh (x) e^{x}$ .

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Step 9.1.2.1

Factor $e^{x}$ out of $e^{x} \cosh (x)$ .

$\frac{e^{x} (\cosh (x)) - \sinh (x) e^{x}}{0 \cdot e^{0 + 2 x} + e^{0 + 2 x}}$

Step 9.1.2.2

Factor $e^{x}$ out of $- \sinh (x) e^{x}$ .

$\frac{e^{x} (\cosh (x)) + e^{x} (- \sinh (x))}{0 \cdot e^{0 + 2 x} + e^{0 + 2 x}}$

Step 9.1.2.3

Factor $e^{x}$ out of $e^{x} (\cosh (x)) + e^{x} (- \sinh (x))$ .

$\frac{e^{x} (\cosh (x) - \sinh (x))}{0 \cdot e^{0 + 2 x} + e^{0 + 2 x}}$

Step 9.2

Simplify the denominator.

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Step 9.2.1

Add $0$ and $2 x$ .

$\frac{e^{x} (\cosh (x) - \sinh (x))}{0 \cdot e^{2 x} + e^{0 + 2 x}}$

Step 9.2.2

Multiply $0$ by $e^{2 x}$ .

$\frac{e^{x} (\cosh (x) - \sinh (x))}{0 + e^{0 + 2 x}}$

Step 9.2.3

Add $0$ and $2 x$ .

$\frac{e^{x} (\cosh (x) - \sinh (x))}{0 + e^{2 x}}$

Step 9.2.4

Add $0$ and $e^{2 x}$ .

$\frac{e^{x} (\cosh (x) - \sinh (x))}{e^{2 x}}$

Step 9.3

Cancel the common factor of $e^{x}$ and $e^{2 x}$ .

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Step 9.3.1

Factor $e^{2 x}$ out of $e^{x} (\cosh (x) - \sinh (x))$ .

$\frac{e^{2 x} (e^{- x} (\cosh (x) - \sinh (x)))}{e^{2 x}}$

Step 9.3.2

Cancel the common factors.

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Step 9.3.2.1

Multiply by $1$ .

$\frac{e^{2 x} (e^{- x} (\cosh (x) - \sinh (x)))}{e^{2 x} \cdot 1}$

Step 9.3.2.2

Cancel the common factor.

$\frac{e^{2 x} (e^{- x} (\cosh (x) - \sinh (x)))}{e^{2 x} \cdot 1}$

Step 9.3.2.3

Rewrite the expression.

$\frac{e^{- x} (\cosh (x) - \sinh (x))}{1}$

Step 9.3.2.4

Divide $e^{- x} (\cosh (x) - \sinh (x))$ by $1$ .

$e^{- x} (\cosh (x) - \sinh (x))$

Step 9.4

Apply the distributive property.

$e^{- x} \cosh (x) + e^{- x} (- \sinh (x))$

Step 9.5

Rewrite using the commutative property of multiplication.

$e^{- x} \cosh (x) - e^{- x} \sinh (x)$

Step 10