Calculus Examples

$c = - 2 + \frac{1000}{q}$ , $q = 25$

Step 1

Differentiate.

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Step 1.1

By the Sum Rule, the derivative of $- 2 + \frac{1000}{q}$ with respect to $q$ is $\frac{d}{d q} [- 2] + \frac{d}{d q} [\frac{1000}{q}]$ .

$\frac{d}{d q} [- 2] + \frac{d}{d q} [\frac{1000}{q}]$

Step 1.2

Since $- 2$ is constant with respect to $q$ , the derivative of $- 2$ with respect to $q$ is $0$ .

$0 + \frac{d}{d q} [\frac{1000}{q}]$

Step 2

Evaluate $\frac{d}{d q} [\frac{1000}{q}]$ .

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Step 2.1

Since $1000$ is constant with respect to $q$ , the derivative of $\frac{1000}{q}$ with respect to $q$ is $1000 \frac{d}{d q} [\frac{1}{q}]$ .

$0 + 1000 \frac{d}{d q} [\frac{1}{q}]$

Step 2.2

Rewrite $\frac{1}{q}$ as $q^{- 1}$ .

$0 + 1000 \frac{d}{d q} [q^{- 1}]$

Step 2.3

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

$0 + 1000 (- q^{- 2})$

Step 2.4

Multiply $- 1$ by $1000$ .

$0 - 1000 q^{- 2}$

Step 3

Simplify.

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Step 3.1

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

$0 - 1000 \frac{1}{q^{2}}$

Step 3.2

Combine terms.

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Step 3.2.1

Combine $- 1000$ and $\frac{1}{q^{2}}$ .

$0 + \frac{- 1000}{q^{2}}$

Step 3.2.2

Move the negative in front of the fraction.

$0 - \frac{1000}{q^{2}}$

Step 3.2.3

Subtract $\frac{1000}{q^{2}}$ from $0$ .

$- \frac{1000}{q^{2}}$

Step 4

Evaluate the derivative at $q = 25$ .

$- \frac{1000}{{(25)}^{2}}$

Step 5

Simplify.

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Step 5.1

Raise $25$ to the power of $2$ .

$- \frac{1000}{625}$

Step 5.2

Cancel the common factor of $1000$ and $625$ .

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Step 5.2.1

Factor $125$ out of $1000$ .

$- \frac{125 (8)}{625}$

Step 5.2.2

Cancel the common factors.

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Step 5.2.2.1

Factor $125$ out of $625$ .

$- \frac{125 \cdot 8}{125 \cdot 5}$

Step 5.2.2.2

Cancel the common factor.

$- \frac{125 \cdot 8}{125 \cdot 5}$

Step 5.2.2.3

Rewrite the expression.

$- \frac{8}{5}$