Calculus Examples

$c (x) = 15000 + 30 x e^{\frac{x}{800}}$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $15000 + 30 x e^{\frac{x}{800}}$ with respect to $x$ is $\frac{d}{d x} [15000] + \frac{d}{d x} [30 x e^{\frac{x}{800}}]$ .

$\frac{d}{d x} [15000] + \frac{d}{d x} [30 x e^{\frac{x}{800}}]$

Step 1.2

Since $15000$ is constant with respect to $x$ , the derivative of $15000$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [30 x e^{\frac{x}{800}}]$

Step 2

Evaluate $\frac{d}{d x} [30 x e^{\frac{x}{800}}]$ .

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Step 2.1

Since $30$ is constant with respect to $x$ , the derivative of $30 x e^{\frac{x}{800}}$ with respect to $x$ is $30 \frac{d}{d x} [x e^{\frac{x}{800}}]$ .

$0 + 30 \frac{d}{d x} [x e^{\frac{x}{800}}]$

Step 2.2

Differentiate using the Product Rule which states that $\frac{d}{d x} [f (x) g (x)]$ is $f (x) \frac{d}{d x} [g (x)] + g (x) \frac{d}{d x} [f (x)]$ where $f (x) = x$ and $g (x) = e^{\frac{x}{800}}$ .

$0 + 30 (x \frac{d}{d x} [e^{\frac{x}{800}}] + e^{\frac{x}{800}} \frac{d}{d x} [x])$

Step 2.3

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 $g (x) = \frac{x}{800}$ .

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Step 2.3.1

To apply the Chain Rule, set $u$ as $\frac{x}{800}$ .

$0 + 30 (x (\frac{d}{d u} [e^{u}] \frac{d}{d x} [\frac{x}{800}]) + e^{\frac{x}{800}} \frac{d}{d x} [x])$

Step 2.3.2

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

$0 + 30 (x (e^{u} \frac{d}{d x} [\frac{x}{800}]) + e^{\frac{x}{800}} \frac{d}{d x} [x])$

Step 2.3.3

Replace all occurrences of $u$ with $\frac{x}{800}$ .

$0 + 30 (x (e^{\frac{x}{800}} \frac{d}{d x} [\frac{x}{800}]) + e^{\frac{x}{800}} \frac{d}{d x} [x])$

Step 2.4

Since $\frac{1}{800}$ is constant with respect to $x$ , the derivative of $\frac{x}{800}$ with respect to $x$ is $\frac{1}{800} \frac{d}{d x} [x]$ .

$0 + 30 (x (e^{\frac{x}{800}} (\frac{1}{800} \frac{d}{d x} [x])) + e^{\frac{x}{800}} \frac{d}{d x} [x])$

Step 2.5

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

$0 + 30 (x (e^{\frac{x}{800}} (\frac{1}{800} \cdot 1)) + e^{\frac{x}{800}} \frac{d}{d x} [x])$

Step 2.6

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

$0 + 30 (x (e^{\frac{x}{800}} (\frac{1}{800} \cdot 1)) + e^{\frac{x}{800}} \cdot 1)$

Step 2.7

Multiply $\frac{1}{800}$ by $1$ .

$0 + 30 (x (e^{\frac{x}{800}} \frac{1}{800}) + e^{\frac{x}{800}} \cdot 1)$

Step 2.8

Combine $e^{\frac{x}{800}}$ and $\frac{1}{800}$ .

$0 + 30 (x \frac{e^{\frac{x}{800}}}{800} + e^{\frac{x}{800}} \cdot 1)$

Step 2.9

Combine $x$ and $\frac{e^{\frac{x}{800}}}{800}$ .

$0 + 30 (\frac{x e^{\frac{x}{800}}}{800} + e^{\frac{x}{800}} \cdot 1)$

Step 2.10

Multiply $e^{\frac{x}{800}}$ by $1$ .

$0 + 30 (\frac{x e^{\frac{x}{800}}}{800} + e^{\frac{x}{800}})$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$0 + 30 \frac{x e^{\frac{x}{800}}}{800} + 30 e^{\frac{x}{800}}$

Step 3.2

Combine terms.

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Step 3.2.1

Combine $30$ and $\frac{x e^{\frac{x}{800}}}{800}$ .

$0 + \frac{30 (x e^{\frac{x}{800}})}{800} + 30 e^{\frac{x}{800}}$

Step 3.2.2

Cancel the common factor of $30$ and $800$ .

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Step 3.2.2.1

Factor $10$ out of $30 x e^{\frac{x}{800}}$ .

$0 + \frac{10 (3 x e^{\frac{x}{800}})}{800} + 30 e^{\frac{x}{800}}$

Step 3.2.2.2

Cancel the common factors.

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Step 3.2.2.2.1

Factor $10$ out of $800$ .

$0 + \frac{10 (3 x e^{\frac{x}{800}})}{10 (80)} + 30 e^{\frac{x}{800}}$

Step 3.2.2.2.2

Cancel the common factor.

$0 + \frac{10 (3 x e^{\frac{x}{800}})}{10 \cdot 80} + 30 e^{\frac{x}{800}}$

Step 3.2.2.2.3

Rewrite the expression.

$0 + \frac{3 x e^{\frac{x}{800}}}{80} + 30 e^{\frac{x}{800}}$

Step 3.2.3

Add $0$ and $\frac{3 x e^{\frac{x}{800}}}{80} + 30 e^{\frac{x}{800}}$ .

$\frac{3 x e^{\frac{x}{800}}}{80} + 30 e^{\frac{x}{800}}$