Calculus Examples

$e^{4 x} + e^{- x}$

Step 1

Write $e^{4 x} + e^{- x}$ as a function.

$f (x) = e^{4 x} + e^{- x}$

Step 2

Find the $x$ values where the second derivative is equal to $0$ .

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Step 2.1

Find the second derivative.

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Step 2.1.1

Find the first derivative.

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Step 2.1.1.1

By the Sum Rule, the derivative of $e^{4 x} + e^{- x}$ with respect to $x$ is $\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [e^{- x}]$ .

$\frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2

Evaluate $\frac{d}{d x} [e^{4 x}]$ .

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Step 2.1.1.2.1

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

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Step 2.1.1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2.1.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2.1.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$e^{4 x} \frac{d}{d x} [4 x] + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2.2

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

$e^{4 x} (4 \frac{d}{d x} [x]) + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2.3

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

$e^{4 x} (4 \cdot 1) + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2.4

Multiply $4$ by $1$ .

$e^{4 x} \cdot 4 + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.2.5

Move $4$ to the left of $e^{4 x}$ .

$4 e^{4 x} + \frac{d}{d x} [e^{- x}]$

Step 2.1.1.3

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

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Step 2.1.1.3.1

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

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Step 2.1.1.3.1.1

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

$4 e^{4 x} + \frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x]$

Step 2.1.1.3.1.2

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

$4 e^{4 x} + e^{u_{2}} \frac{d}{d x} [- x]$

Step 2.1.1.3.1.3

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

$4 e^{4 x} + e^{- x} \frac{d}{d x} [- x]$

Step 2.1.1.3.2

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$4 e^{4 x} + e^{- x} (- \frac{d}{d x} [x])$

Step 2.1.1.3.3

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

$4 e^{4 x} + e^{- x} (- 1 \cdot 1)$

Step 2.1.1.3.4

Multiply $- 1$ by $1$ .

$4 e^{4 x} + e^{- x} \cdot - 1$

Step 2.1.1.3.5

Move $- 1$ to the left of $e^{- x}$ .

$4 e^{4 x} - 1 \cdot e^{- x}$

Step 2.1.1.3.6

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$f' (x) = 4 e^{4 x} - e^{- x}$

Step 2.1.2

Find the second derivative.

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Step 2.1.2.1

By the Sum Rule, the derivative of $4 e^{4 x} - e^{- x}$ with respect to $x$ is $\frac{d}{d x} [4 e^{4 x}] + \frac{d}{d x} [- e^{- x}]$ .

$\frac{d}{d x} [4 e^{4 x}] + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2

Evaluate $\frac{d}{d x} [4 e^{4 x}]$ .

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Step 2.1.2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 e^{4 x}$ with respect to $x$ is $4 \frac{d}{d x} [e^{4 x}]$ .

$4 \frac{d}{d x} [e^{4 x}] + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.2

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

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Step 2.1.2.2.2.1

To apply the Chain Rule, set $u_{1}$ as $4 x$ .

$4 (\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [4 x]) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.2.2

Differentiate using the Exponential Rule which states that $\frac{d}{d u_{1}} [a^{u_{1}}]$ is $a^{u_{1}} \ln (a)$ where $a$ = $e$ .

$4 (e^{u_{1}} \frac{d}{d x} [4 x]) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.2.3

Replace all occurrences of $u_{1}$ with $4 x$ .

$4 (e^{4 x} \frac{d}{d x} [4 x]) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.3

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

$4 (e^{4 x} (4 \frac{d}{d x} [x])) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.4

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

$4 (e^{4 x} (4 \cdot 1)) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.5

Multiply $4$ by $1$ .

$4 (e^{4 x} \cdot 4) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.6

Move $4$ to the left of $e^{4 x}$ .

$4 (4 \cdot e^{4 x}) + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.2.7

Multiply $4$ by $4$ .

$16 e^{4 x} + \frac{d}{d x} [- e^{- x}]$

Step 2.1.2.3

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

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Step 2.1.2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- e^{- x}$ with respect to $x$ is $- \frac{d}{d x} [e^{- x}]$ .

$16 e^{4 x} - \frac{d}{d x} [e^{- x}]$

Step 2.1.2.3.2

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

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Step 2.1.2.3.2.1

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

$16 e^{4 x} - (\frac{d}{d u_{2}} [e^{u_{2}}] \frac{d}{d x} [- x])$

Step 2.1.2.3.2.2

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

$16 e^{4 x} - (e^{u_{2}} \frac{d}{d x} [- x])$

Step 2.1.2.3.2.3

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

$16 e^{4 x} - (e^{- x} \frac{d}{d x} [- x])$

Step 2.1.2.3.3

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$16 e^{4 x} - (e^{- x} (- \frac{d}{d x} [x]))$

Step 2.1.2.3.4

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

$16 e^{4 x} - (e^{- x} (- 1 \cdot 1))$

Step 2.1.2.3.5

Multiply $- 1$ by $1$ .

$16 e^{4 x} - (e^{- x} \cdot - 1)$

Step 2.1.2.3.6

Move $- 1$ to the left of $e^{- x}$ .

$16 e^{4 x} - (- 1 \cdot e^{- x})$

Step 2.1.2.3.7

Rewrite $- 1 e^{- x}$ as $- e^{- x}$ .

$16 e^{4 x} - - e^{- x}$

Step 2.1.2.3.8

Multiply $- 1$ by $- 1$ .

$16 e^{4 x} + 1 e^{- x}$

Step 2.1.2.3.9

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

$f'' (x) = 16 e^{4 x} + e^{- x}$

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $16 e^{4 x} + e^{- x}$ .

$16 e^{4 x} + e^{- x}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $16 e^{4 x} + e^{- x} = 0$ .

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Step 2.2.1

Set the second derivative equal to $0$ .

$16 e^{4 x} + e^{- x} = 0$

Step 2.2.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

Step 3

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Interval Notation:

$(- \infty, \infty)$

Set-Builder Notation:

${x | x \in ℝ}$

Step 4

The graph is concave up because the second derivative is positive.

The graph is concave up

Step 5