Calculus Examples

$f (x) = 15 + 3 e^{- 3 x}$

Step 1

Find the first derivative of the function.

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Step 1.1

Differentiate.

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Step 1.1.1

By the Sum Rule, the derivative of $15 + 3 e^{- 3 x}$ with respect to $x$ is $\frac{d}{d x} [15] + \frac{d}{d x} [3 e^{- 3 x}]$ .

$\frac{d}{d x} [15] + \frac{d}{d x} [3 e^{- 3 x}]$

Step 1.1.2

Since $15$ is constant with respect to $x$ , the derivative of $15$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [3 e^{- 3 x}]$

Step 1.2

Evaluate $\frac{d}{d x} [3 e^{- 3 x}]$ .

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Step 1.2.1

Since $3$ is constant with respect to $x$ , the derivative of $3 e^{- 3 x}$ with respect to $x$ is $3 \frac{d}{d x} [e^{- 3 x}]$ .

$0 + 3 \frac{d}{d x} [e^{- 3 x}]$

Step 1.2.2

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 $g (x) = - 3 x$ .

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Step 1.2.2.1

To apply the Chain Rule, set $u$ as $- 3 x$ .

$0 + 3 (\frac{d}{d u} [e^{u}] \frac{d}{d x} [- 3 x])$

Step 1.2.2.2

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

$0 + 3 (e^{u} \frac{d}{d x} [- 3 x])$

Step 1.2.2.3

Replace all occurrences of $u$ with $- 3 x$ .

$0 + 3 (e^{- 3 x} \frac{d}{d x} [- 3 x])$

Step 1.2.3

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$0 + 3 (e^{- 3 x} (- 3 \frac{d}{d x} [x]))$

Step 1.2.4

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

$0 + 3 (e^{- 3 x} (- 3 \cdot 1))$

Step 1.2.5

Multiply $- 3$ by $1$ .

$0 + 3 (e^{- 3 x} \cdot - 3)$

Step 1.2.6

Move $- 3$ to the left of $e^{- 3 x}$ .

$0 + 3 (- 3 \cdot e^{- 3 x})$

Step 1.2.7

Multiply $- 3$ by $3$ .

$0 - 9 e^{- 3 x}$

Step 1.3

Subtract $9 e^{- 3 x}$ from $0$ .

$- 9 e^{- 3 x}$

Step 2

Find the second derivative of the function.

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Step 2.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 e^{- 3 x}$ with respect to $x$ is $- 9 \frac{d}{d x} [e^{- 3 x}]$ .

$f'' (x) = - 9 \frac{d}{d x} e^{- 3 x}$

Step 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 $g (x) = - 3 x$ .

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Step 2.2.1

To apply the Chain Rule, set $u$ as $- 3 x$ .

$f'' (x) = - 9 (\frac{d}{d u} (e^{u}) \frac{d}{d x} (- 3 x))$

Step 2.2.2

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

$f'' (x) = - 9 (e^{u} \frac{d}{d x} (- 3 x))$

Step 2.2.3

Replace all occurrences of $u$ with $- 3 x$ .

$f'' (x) = - 9 (e^{- 3 x} \frac{d}{d x} (- 3 x))$

Step 2.3

Differentiate.

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Step 2.3.1

Since $- 3$ is constant with respect to $x$ , the derivative of $- 3 x$ with respect to $x$ is $- 3 \frac{d}{d x} [x]$ .

$f'' (x) = - 9 e^{- 3 x} (- 3 \frac{d}{d x} x)$

Step 2.3.2

Multiply $- 3$ by $- 9$ .

$f'' (x) = 27 e^{- 3 x} \frac{d}{d x} (x)$

Step 2.3.3

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

$f'' (x) = 27 e^{- 3 x} \cdot 1$

Step 2.3.4

Multiply $27$ by $1$ .

$f'' (x) = 27 e^{- 3 x}$

Step 3

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$- 9 e^{- 3 x} = 0$

Step 4

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 5

No Local Extrema

Step 6