Calculus Examples

$e^{2 x} + e^{- x}$

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

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

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

Step 1.1.2

Evaluate $\frac{d}{d x} [e^{2 x}]$ .

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Step 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 $g (x) = 2 x$ .

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Step 1.1.2.1.1

To apply the Chain Rule, set $u_{1}$ as $2 x$ .

$\frac{d}{d u_{1}} [e^{u_{1}}] \frac{d}{d x} [2 x] + \frac{d}{d x} [e^{- x}]$

Step 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 $a$ = $e$ .

$e^{u_{1}} \frac{d}{d x} [2 x] + \frac{d}{d x} [e^{- x}]$

Step 1.1.2.1.3

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Step 1.1.2.2

Since $2$ is constant with respect to $x$ , the derivative of $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 1.1.2.3

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

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

Step 1.1.2.4

Multiply $2$ by $1$ .

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

Step 1.1.2.5

Move $2$ to the left of $e^{2 x}$ .

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

Step 1.1.3

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

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Step 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 1.1.3.1.1

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

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

Step 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 $a$ = $e$ .

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

Step 1.1.3.1.3

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

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

Step 1.1.3.2

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

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

Step 1.1.3.3

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

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

Step 1.1.3.4

Multiply $- 1$ by $1$ .

$2 e^{2 x} + e^{- x} \cdot - 1$

Step 1.1.3.5

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

$2 e^{2 x} - 1 \cdot e^{- x}$

Step 1.1.3.6

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

$f' (x) = 2 e^{2 x} - e^{- x}$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $2 e^{2 x} - e^{- x}$ .

$2 e^{2 x} - e^{- x}$

Step 2

Set the first derivative equal to $0$ then solve the equation $2 e^{2 x} - e^{- x} = 0$ .

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Step 2.1

Set the first derivative equal to $0$ .

$2 e^{2 x} - e^{- x} = 0$

Step 2.2

Move $- e^{- x}$ to the right side of the equation by adding it to both sides.

$2 e^{2 x} = e^{- x}$

Step 2.3

Take the natural logarithm of both sides of the equation to remove the variable from the exponent.

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

Step 2.4

Expand the left side.

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Step 2.4.1

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

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

Step 2.4.2

Expand $\ln (e^{2 x})$ by moving $2 x$ outside the logarithm.

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

Step 2.4.3

The natural logarithm of $e$ is $1$ .

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

Step 2.4.4

Multiply $2$ by $1$ .

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

Step 2.5

Expand the right side.

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Step 2.5.1

Expand $\ln (e^{- x})$ by moving $- x$ outside the logarithm.

$\ln (2) + 2 x = - x \ln (e)$

Step 2.5.2

The natural logarithm of $e$ is $1$ .

$\ln (2) + 2 x = - x \cdot 1$

Step 2.5.3

Multiply $- 1$ by $1$ .

$\ln (2) + 2 x = - x$

Step 2.6

Move all terms containing $x$ to the left side of the equation.

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Step 2.6.1

Add $x$ to both sides of the equation.

$\ln (2) + 2 x + x = 0$

Step 2.6.2

Add $2 x$ and $x$ .

$\ln (2) + 3 x = 0$

Step 2.7

Subtract $\ln (2)$ from both sides of the equation.

$3 x = - \ln (2)$

Step 2.8

Divide each term in $3 x = - \ln (2)$ by $3$ and simplify.

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Step 2.8.1

Divide each term in $3 x = - \ln (2)$ by $3$ .

$\frac{3 x}{3} = \frac{- \ln (2)}{3}$

Step 2.8.2

Simplify the left side.

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Step 2.8.2.1

Cancel the common factor of $3$ .

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Step 2.8.2.1.1

Cancel the common factor.

$\frac{3 x}{3} = \frac{- \ln (2)}{3}$

Step 2.8.2.1.2

Divide $x$ by $1$ .

$x = \frac{- \ln (2)}{3}$

Step 2.8.3

Simplify the right side.

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Step 2.8.3.1

Move the negative in front of the fraction.

$x = - \frac{\ln (2)}{3}$

Step 3

Find the values where the derivative is undefined.

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Step 3.1

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.

Step 4

Evaluate $e^{2 x} + e^{- x}$ at each $x$ value where the derivative is $0$ or undefined.

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Step 4.1

Evaluate at $x = - \frac{\ln (2)}{3}$ .

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Step 4.1.1

Substitute $- \frac{\ln (2)}{3}$ for $x$ .

$e^{2 (- \frac{\ln (2)}{3})} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2

Simplify each term.

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Step 4.1.2.1

Rewrite $\frac{\ln (2)}{3}$ as $\frac{1}{3} \ln (2)$ .

$e^{2 (- (\frac{1}{3} \ln (2)))} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.2

Simplify $\frac{1}{3} \ln (2)$ by moving $\frac{1}{3}$ inside the logarithm.

$e^{2 (- \ln (2^{\frac{1}{3}}))} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.3

Multiply $2 (- \ln (2^{\frac{1}{3}}))$ .

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Step 4.1.2.3.1

Multiply $- 1$ by $2$ .

$e^{- 2 \ln (2^{\frac{1}{3}})} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.3.2

Simplify $- 2 \ln (2^{\frac{1}{3}})$ by moving $2$ inside the logarithm.

$e^{- \ln ({(2^{\frac{1}{3}})}^{2})} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.4

Simplify $- \ln ({(2^{\frac{1}{3}})}^{2})$ by moving $- 1$ inside the logarithm.

$e^{\ln ({({(2^{\frac{1}{3}})}^{2})}^{- 1})} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.5

Exponentiation and log are inverse functions.

${({(2^{\frac{1}{3}})}^{2})}^{- 1} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.6

Multiply the exponents in ${({(2^{\frac{1}{3}})}^{2})}^{- 1}$ .

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Step 4.1.2.6.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.1.2.6.2

Multiply $2$ by $- 1$ .

${(2^{\frac{1}{3}})}^{- 2} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.7

Multiply the exponents in ${(2^{\frac{1}{3}})}^{- 2}$ .

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Step 4.1.2.7.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.1.2.7.2

Combine $\frac{1}{3}$ and $- 2$ .

$2^{\frac{- 2}{3}} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.7.3

Move the negative in front of the fraction.

$2^{- \frac{2}{3}} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.8

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2^{\frac{2}{3}}} + e^{- (- \frac{\ln (2)}{3})}$

Step 4.1.2.9

Rewrite $\frac{\ln (2)}{3}$ as $\frac{1}{3} \ln (2)$ .

$\frac{1}{2^{\frac{2}{3}}} + e^{- - (\frac{1}{3} \ln (2))}$

Step 4.1.2.10

Simplify $\frac{1}{3} \ln (2)$ by moving $\frac{1}{3}$ inside the logarithm.

$\frac{1}{2^{\frac{2}{3}}} + e^{- - \ln (2^{\frac{1}{3}})}$

Step 4.1.2.11

Multiply $- - \ln (2^{\frac{1}{3}})$ .

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Step 4.1.2.11.1

Multiply $- 1$ by $- 1$ .

$\frac{1}{2^{\frac{2}{3}}} + e^{1 \ln (2^{\frac{1}{3}})}$

Step 4.1.2.11.2

Multiply $\ln (2^{\frac{1}{3}})$ by $1$ .

$\frac{1}{2^{\frac{2}{3}}} + e^{\ln (2^{\frac{1}{3}})}$

Step 4.1.2.12

Exponentiation and log are inverse functions.

$\frac{1}{2^{\frac{2}{3}}} + 2^{\frac{1}{3}}$

Step 4.2

List all of the points.

$(- \frac{\ln (2)}{3}, \frac{1}{2^{\frac{2}{3}}} + 2^{\frac{1}{3}})$

Step 5