Calculus Examples

$e^{- 5 x}$

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.

$f (x + h) = e^{- 5 (x + h)}$

Step 2.1.2

Simplify the result.

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Step 2.1.2.1

Apply the distributive property.

$f (x + h) = e^{- 5 x - 5 h}$

Step 2.1.2.2

The final answer is $e^{- 5 x - 5 h}$ .

$e^{- 5 x - 5 h}$

Step 2.2

Find the components of the definition.

$f (x + h) = e^{- 5 x - 5 h}$

$f (x) = e^{- 5 x}$

$f (x + h) = e^{- 5 x - 5 h}$

$f (x) = e^{- 5 x}$

Step 3

Plug in the components.

$f' (x) = lim_{h \to 0} \frac{e^{- 5 x - 5 h} - (e^{- 5 x})}{h}$

Step 4

Multiply $- 1$ by $e^{- 5 x}$ .

$f' (x) = lim_{h \to 0} \frac{e^{- 5 x - 5 h} - e^{- 5 x}}{h}$

Step 5

Apply L'Hospital's rule.

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Step 5.1

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

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Step 5.1.1

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

$\frac{lim_{h \to 0} e^{- 5 x - 5 h} - e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2

Evaluate the limit of the numerator.

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Step 5.1.2.1

Evaluate the limit.

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Step 5.1.2.1.1

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

$\frac{lim_{h \to 0} e^{- 5 x - 5 h} - lim_{h \to 0} e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.1.2

Move the limit into the exponent.

$\frac{e^{lim_{h \to 0} - 5 x - 5 h} - lim_{h \to 0} e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.1.3

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

$\frac{e^{- lim_{h \to 0} 5 x - lim_{h \to 0} 5 h} - lim_{h \to 0} e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.1.4

Evaluate the limit of $5 x$ which is constant as $h$ approaches $0$ .

$\frac{e^{- 5 x - lim_{h \to 0} 5 h} - lim_{h \to 0} e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.1.5

Move the term $5$ outside of the limit because it is constant with respect to $h$ .

$\frac{e^{- 5 x - 5 lim_{h \to 0} h} - lim_{h \to 0} e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.1.6

Evaluate the limit of $e^{- 5 x}$ which is constant as $h$ approaches $0$ .

$\frac{e^{- 5 x - 5 lim_{h \to 0} h} - e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.2

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

$\frac{e^{- 5 x - 5 \cdot 0} - e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.3

Simplify the answer.

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Step 5.1.2.3.1

Simplify each term.

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Step 5.1.2.3.1.1

Multiply $- 5$ by $0$ .

$\frac{e^{- 5 x + 0} - e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.3.1.2

Add $- 5 x$ and $0$ .

$\frac{e^{- 5 x} - e^{- 5 x}}{lim_{h \to 0} h}$

Step 5.1.2.3.2

Subtract $e^{- 5 x}$ from $e^{- 5 x}$ .

$\frac{0}{lim_{h \to 0} h}$

Step 5.1.3

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

$\frac{0}{0}$

Step 5.1.4

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

Undefined

$\frac{0}{0}$

Step 5.2

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.

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} - e^{- 5 x}}{h} = lim_{h \to 0} \frac{\frac{d}{d h} [e^{- 5 x - 5 h} - e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

$lim_{h \to 0} \frac{\frac{d}{d h} [e^{- 5 x - 5 h} - e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.2

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

$lim_{h \to 0} \frac{\frac{d}{d h} [e^{- 5 x - 5 h}] + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3

Evaluate $\frac{d}{d h} [e^{- 5 x - 5 h}]$ .

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Step 5.3.3.1

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 $g (h) = - 5 x - 5 h$ .

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Step 5.3.3.1.1

To apply the Chain Rule, set $u$ as $- 5 x - 5 h$ .

$lim_{h \to 0} \frac{\frac{d}{d u} [e^{u}] \frac{d}{d h} [- 5 x - 5 h] + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.1.2

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

$lim_{h \to 0} \frac{e^{u} \frac{d}{d h} [- 5 x - 5 h] + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.1.3

Replace all occurrences of $u$ with $- 5 x - 5 h$ .

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} \frac{d}{d h} [- 5 x - 5 h] + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.2

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

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} (\frac{d}{d h} [- 5 x] + \frac{d}{d h} [- 5 h]) + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.3

Since $- 5 x$ is constant with respect to $h$ , the derivative of $- 5 x$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} (0 + \frac{d}{d h} [- 5 h]) + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.4

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

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} (0 - 5 \frac{d}{d h} [h]) + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.5

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

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} (0 - 5 \cdot 1) + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.6

Multiply $- 5$ by $1$ .

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} (0 - 5) + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.7

Subtract $5$ from $0$ .

$lim_{h \to 0} \frac{e^{- 5 x - 5 h} \cdot - 5 + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.3.8

Move $- 5$ to the left of $e^{- 5 x - 5 h}$ .

$lim_{h \to 0} \frac{- 5 e^{- 5 x - 5 h} + \frac{d}{d h} [- e^{- 5 x}]}{\frac{d}{d h} [h]}$

Step 5.3.4

Since $- e^{- 5 x}$ is constant with respect to $h$ , the derivative of $- e^{- 5 x}$ with respect to $h$ is $0$ .

$lim_{h \to 0} \frac{- 5 e^{- 5 x - 5 h} + 0}{\frac{d}{d h} [h]}$

Step 5.3.5

Add $- 5 e^{- 5 x - 5 h}$ and $0$ .

$lim_{h \to 0} \frac{- 5 e^{- 5 x - 5 h}}{\frac{d}{d h} [h]}$

Step 5.3.6

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

$lim_{h \to 0} \frac{- 5 e^{- 5 x - 5 h}}{1}$

Step 5.4

Divide $- 5 e^{- 5 x - 5 h}$ by $1$ .

$lim_{h \to 0} - 5 e^{- 5 x - 5 h}$

Step 6

Evaluate the limit.

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Step 6.1

Move the term $- 5$ outside of the limit because it is constant with respect to $h$ .

$- 5 lim_{h \to 0} e^{- 5 x - 5 h}$

Step 6.2

Move the limit into the exponent.

$- 5 e^{lim_{h \to 0} - 5 x - 5 h}$

Step 6.3

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

$- 5 e^{- lim_{h \to 0} 5 x - lim_{h \to 0} 5 h}$

Step 6.4

Evaluate the limit of $5 x$ which is constant as $h$ approaches $0$ .

$- 5 e^{- 5 x - lim_{h \to 0} 5 h}$

Step 6.5

Move the term $5$ outside of the limit because it is constant with respect to $h$ .

$- 5 e^{- 5 x - 5 lim_{h \to 0} h}$

Step 7

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

$- 5 e^{- 5 x - 5 \cdot 0}$

Step 8

Simplify the answer.

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Step 8.1

Multiply $- 5$ by $0$ .

$- 5 e^{- 5 x + 0}$

Step 8.2

Add $- 5 x$ and $0$ .

$- 5 e^{- 5 x}$

Step 9