Calculus Examples

$e^{x}$

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.

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

Step 2.1.2

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

$e^{x + h}$

Step 2.2

Find the components of the definition.

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

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

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

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

Step 3

Plug in the components.

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

Step 4

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

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

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.

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

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 $0$ .

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

Step 5.1.2.1.2

Move the limit into the exponent.

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

Step 5.1.2.1.3

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

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

Step 5.1.2.1.4

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

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

Step 5.1.2.1.5

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

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

Step 5.1.2.2

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

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

Step 5.1.2.3

Combine the opposite terms in $e^{x + 0} - e^{x}$ .

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Step 5.1.2.3.1

Add $x$ and $0$ .

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

Step 5.1.2.3.2

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

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

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.

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

Step 5.3

Find the derivative of the numerator and denominator.

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Step 5.3.1

Differentiate the numerator and denominator.

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

Step 5.3.2

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

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

Step 5.3.3

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

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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 $g (h) = x + h$ .

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Step 5.3.3.1.1

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

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

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 $a$ = $e$ .

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

Step 5.3.3.1.3

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

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

Step 5.3.3.2

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

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

Step 5.3.3.3

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

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

Step 5.3.3.4

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

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

Step 5.3.3.5

Add $0$ and $1$ .

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

Step 5.3.3.6

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

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

Step 5.3.4

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

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

Step 5.3.5

Add $e^{x + h}$ and $0$ .

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

Step 5.3.6

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

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

Step 5.4

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

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

Step 6

Evaluate the limit.

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Step 6.1

Move the limit into the exponent.

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

Step 6.2

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

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

Step 6.3

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

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

Step 7

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

$e^{x + 0}$

Step 8

Add $x$ and $0$ .

$e^{x}$

Step 9