Calculus Examples

$\frac{1000}{2 t + 6} - \frac{2000}{{(2 t + 6)}^{2}}$

Step 1

By the Sum Rule, the derivative of $\frac{1000}{2 t + 6} - \frac{2000}{{(2 t + 6)}^{2}}$ with respect to $t$ is $\frac{d}{d t} [\frac{1000}{2 t + 6}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$ .

$\frac{d}{d t} [\frac{1000}{2 t + 6}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2

Evaluate $\frac{d}{d t} [\frac{1000}{2 t + 6}]$ .

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Step 2.1

Cancel the common factor of $1000$ and $2 t + 6$ .

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Step 2.1.1

Factor $2$ out of $1000$ .

$\frac{d}{d t} [\frac{2 \cdot 500}{2 t + 6}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.1.2

Cancel the common factors.

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Step 2.1.2.1

Factor $2$ out of $2 t$ .

$\frac{d}{d t} [\frac{2 \cdot 500}{2 (t) + 6}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.1.2.2

Factor $2$ out of $6$ .

$\frac{d}{d t} [\frac{2 \cdot 500}{2 (t) + 2 (3)}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.1.2.3

Factor $2$ out of $2 (t) + 2 (3)$ .

$\frac{d}{d t} [\frac{2 \cdot 500}{2 (t + 3)}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.1.2.4

Cancel the common factor.

$\frac{d}{d t} [\frac{2 \cdot 500}{2 (t + 3)}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.1.2.5

Rewrite the expression.

$\frac{d}{d t} [\frac{500}{t + 3}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.2

Since $500$ is constant with respect to $t$ , the derivative of $\frac{500}{t + 3}$ with respect to $t$ is $500 \frac{d}{d t} [\frac{1}{t + 3}]$ .

$500 \frac{d}{d t} [\frac{1}{t + 3}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.3

Rewrite $\frac{1}{t + 3}$ as ${(t + 3)}^{- 1}$ .

$500 \frac{d}{d t} [{(t + 3)}^{- 1}] + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = t + 3$ .

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Step 2.4.1

To apply the Chain Rule, set $u_{1}$ as $t + 3$ .

$500 (\frac{d}{d u_{1}} [{u_{1}}^{- 1}] \frac{d}{d t} [t + 3]) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.4.2

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

$500 (- {u_{1}}^{- 2} \frac{d}{d t} [t + 3]) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.4.3

Replace all occurrences of $u_{1}$ with $t + 3$ .

$500 (- {(t + 3)}^{- 2} \frac{d}{d t} [t + 3]) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.5

By the Sum Rule, the derivative of $t + 3$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [3]$ .

$500 (- {(t + 3)}^{- 2} (\frac{d}{d t} [t] + \frac{d}{d t} [3])) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.6

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

$500 (- {(t + 3)}^{- 2} (1 + \frac{d}{d t} [3])) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.7

Since $3$ is constant with respect to $t$ , the derivative of $3$ with respect to $t$ is $0$ .

$500 (- {(t + 3)}^{- 2} (1 + 0)) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.8

Add $1$ and $0$ .

$500 (- {(t + 3)}^{- 2} \cdot 1) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.9

Multiply $- 1$ by $1$ .

$500 (- {(t + 3)}^{- 2}) + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 2.10

Multiply $- 1$ by $500$ .

$- 500 {(t + 3)}^{- 2} + \frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$

Step 3

Evaluate $\frac{d}{d t} [- \frac{2000}{{(2 t + 6)}^{2}}]$ .

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Step 3.1

Since $- 2000$ is constant with respect to $t$ , the derivative of $- \frac{2000}{{(2 t + 6)}^{2}}$ with respect to $t$ is $- 2000 \frac{d}{d t} [\frac{1}{{(2 t + 6)}^{2}}]$ .

$- 500 {(t + 3)}^{- 2} - 2000 \frac{d}{d t} [\frac{1}{{(2 t + 6)}^{2}}]$

Step 3.2

Rewrite $\frac{1}{{(2 t + 6)}^{2}}$ as ${({(2 t + 6)}^{2})}^{- 1}$ .

$- 500 {(t + 3)}^{- 2} - 2000 \frac{d}{d t} [{({(2 t + 6)}^{2})}^{- 1}]$

Step 3.3

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = {(2 t + 6)}^{2}$ .

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Step 3.3.1

To apply the Chain Rule, set $u_{2}$ as ${(2 t + 6)}^{2}$ .

$- 500 {(t + 3)}^{- 2} - 2000 (\frac{d}{d u_{2}} [{u_{2}}^{- 1}] \frac{d}{d t} [{(2 t + 6)}^{2}])$

Step 3.3.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = - 1$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {u_{2}}^{- 2} \frac{d}{d t} [{(2 t + 6)}^{2}])$

Step 3.3.3

Replace all occurrences of $u_{2}$ with ${(2 t + 6)}^{2}$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} \frac{d}{d t} [{(2 t + 6)}^{2}])$

Step 3.4

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{2}$ and $g (t) = 2 t + 6$ .

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Step 3.4.1

To apply the Chain Rule, set $u_{3}$ as $2 t + 6$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (\frac{d}{d u_{3}} [{u_{3}}^{2}] \frac{d}{d t} [2 t + 6]))$

Step 3.4.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{3}} [{u_{3}}^{n}]$ is $n {u_{3}}^{n - 1}$ where $n = 2$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (2 u_{3} \frac{d}{d t} [2 t + 6]))$

Step 3.4.3

Replace all occurrences of $u_{3}$ with $2 t + 6$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (2 (2 t + 6) \frac{d}{d t} [2 t + 6]))$

Step 3.5

By the Sum Rule, the derivative of $2 t + 6$ with respect to $t$ is $\frac{d}{d t} [2 t] + \frac{d}{d t} [6]$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (2 (2 t + 6) (\frac{d}{d t} [2 t] + \frac{d}{d t} [6])))$

Step 3.6

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

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (2 (2 t + 6) (2 \frac{d}{d t} [t] + \frac{d}{d t} [6])))$

Step 3.7

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

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (2 (2 t + 6) (2 \cdot 1 + \frac{d}{d t} [6])))$

Step 3.8

Since $6$ is constant with respect to $t$ , the derivative of $6$ with respect to $t$ is $0$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {({(2 t + 6)}^{2})}^{- 2} (2 (2 t + 6) (2 \cdot 1 + 0)))$

Step 3.9

Multiply the exponents in ${({(2 t + 6)}^{2})}^{- 2}$ .

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Step 3.9.1

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

$- 500 {(t + 3)}^{- 2} - 2000 (- {(2 t + 6)}^{2 \cdot - 2} (2 (2 t + 6) (2 \cdot 1 + 0)))$

Step 3.9.2

Multiply $2$ by $- 2$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {(2 t + 6)}^{- 4} (2 (2 t + 6) (2 \cdot 1 + 0)))$

Step 3.10

Multiply $2$ by $1$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {(2 t + 6)}^{- 4} (2 (2 t + 6) (2 + 0)))$

Step 3.11

Add $2$ and $0$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {(2 t + 6)}^{- 4} (2 (2 t + 6) \cdot 2))$

Step 3.12

Multiply $2$ by $2$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- {(2 t + 6)}^{- 4} (4 (2 t + 6)))$

Step 3.13

Multiply $4$ by $- 1$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- 4 {(2 t + 6)}^{- 4} (2 t + 6))$

Step 3.14

Raise $2 t + 6$ to the power of $1$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- 4 ({(2 t + 6)}^{1} {(2 t + 6)}^{- 4}))$

Step 3.15

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$- 500 {(t + 3)}^{- 2} - 2000 (- 4 {(2 t + 6)}^{1 - 4})$

Step 3.16

Subtract $4$ from $1$ .

$- 500 {(t + 3)}^{- 2} - 2000 (- 4 {(2 t + 6)}^{- 3})$

Step 3.17

Multiply $- 4$ by $- 2000$ .

$- 500 {(t + 3)}^{- 2} + 8000 {(2 t + 6)}^{- 3}$

Step 4

Simplify.

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Step 4.1

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

$- 500 \frac{1}{{(t + 3)}^{2}} + 8000 {(2 t + 6)}^{- 3}$

Step 4.2

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

$- 500 \frac{1}{{(t + 3)}^{2}} + 8000 \frac{1}{{(2 t + 6)}^{3}}$

Step 4.3

Combine terms.

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Step 4.3.1

Combine $- 500$ and $\frac{1}{{(t + 3)}^{2}}$ .

$\frac{- 500}{{(t + 3)}^{2}} + 8000 \frac{1}{{(2 t + 6)}^{3}}$

Step 4.3.2

Move the negative in front of the fraction.

$- \frac{500}{{(t + 3)}^{2}} + 8000 \frac{1}{{(2 t + 6)}^{3}}$

Step 4.3.3

Combine $8000$ and $\frac{1}{{(2 t + 6)}^{3}}$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8000}{{(2 t + 6)}^{3}}$

Step 4.4

Simplify each term.

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Step 4.4.1

Simplify the denominator.

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Step 4.4.1.1

Factor $2$ out of $2 t + 6$ .

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Step 4.4.1.1.1

Factor $2$ out of $2 t$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8000}{{(2 (t) + 6)}^{3}}$

Step 4.4.1.1.2

Factor $2$ out of $6$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8000}{{(2 t + 2 \cdot 3)}^{3}}$

Step 4.4.1.1.3

Factor $2$ out of $2 t + 2 \cdot 3$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8000}{{(2 (t + 3))}^{3}}$

Step 4.4.1.2

Apply the product rule to $2 (t + 3)$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8000}{2^{3} {(t + 3)}^{3}}$

Step 4.4.1.3

Raise $2$ to the power of $3$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8000}{8 {(t + 3)}^{3}}$

Step 4.4.2

Cancel the common factor of $8000$ and $8$ .

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Step 4.4.2.1

Factor $8$ out of $8000$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8 \cdot 1000}{8 {(t + 3)}^{3}}$

Step 4.4.2.2

Cancel the common factors.

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Step 4.4.2.2.1

Factor $8$ out of $8 {(t + 3)}^{3}$ .

$- \frac{500}{{(t + 3)}^{2}} + \frac{8 \cdot 1000}{8 ({(t + 3)}^{3})}$

Step 4.4.2.2.2

Cancel the common factor.

$- \frac{500}{{(t + 3)}^{2}} + \frac{8 \cdot 1000}{8 {(t + 3)}^{3}}$

Step 4.4.2.2.3

Rewrite the expression.

$- \frac{500}{{(t + 3)}^{2}} + \frac{1000}{{(t + 3)}^{3}}$