Calculus Examples

$f (x) = \frac{t - 5}{t^{2} + 10 t + 25}$

Step 1

Differentiate using the Quotient Rule which states that $\frac{d}{d t} [\frac{f (t)}{g (t)}]$ is $\frac{g (t) \frac{d}{d t} [f (t)] - f (t) \frac{d}{d t} [g (t)]}{{g (t)}^{2}}$ where $f (t) = t - 5$ and $g (t) = t^{2} + 10 t + 25$ .

$\frac{(t^{2} + 10 t + 25) \frac{d}{d t} [t - 5] - (t - 5) \frac{d}{d t} [t^{2} + 10 t + 25]}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2

Differentiate.

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Step 2.1

By the Sum Rule, the derivative of $t - 5$ with respect to $t$ is $\frac{d}{d t} [t] + \frac{d}{d t} [- 5]$ .

$\frac{(t^{2} + 10 t + 25) (\frac{d}{d t} [t] + \frac{d}{d t} [- 5]) - (t - 5) \frac{d}{d t} [t^{2} + 10 t + 25]}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.2

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

$\frac{(t^{2} + 10 t + 25) (1 + \frac{d}{d t} [- 5]) - (t - 5) \frac{d}{d t} [t^{2} + 10 t + 25]}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.3

Since $- 5$ is constant with respect to $t$ , the derivative of $- 5$ with respect to $t$ is $0$ .

$\frac{(t^{2} + 10 t + 25) (1 + 0) - (t - 5) \frac{d}{d t} [t^{2} + 10 t + 25]}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.4

Simplify the expression.

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Step 2.4.1

Add $1$ and $0$ .

$\frac{(t^{2} + 10 t + 25) \cdot 1 - (t - 5) \frac{d}{d t} [t^{2} + 10 t + 25]}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.4.2

Multiply $t^{2} + 10 t + 25$ by $1$ .

$\frac{t^{2} + 10 t + 25 - (t - 5) \frac{d}{d t} [t^{2} + 10 t + 25]}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.5

By the Sum Rule, the derivative of $t^{2} + 10 t + 25$ with respect to $t$ is $\frac{d}{d t} [t^{2}] + \frac{d}{d t} [10 t] + \frac{d}{d t} [25]$ .

$\frac{t^{2} + 10 t + 25 - (t - 5) (\frac{d}{d t} [t^{2}] + \frac{d}{d t} [10 t] + \frac{d}{d t} [25])}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.6

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

$\frac{t^{2} + 10 t + 25 - (t - 5) (2 t + \frac{d}{d t} [10 t] + \frac{d}{d t} [25])}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.7

Since $10$ is constant with respect to $t$ , the derivative of $10 t$ with respect to $t$ is $10 \frac{d}{d t} [t]$ .

$\frac{t^{2} + 10 t + 25 - (t - 5) (2 t + 10 \frac{d}{d t} [t] + \frac{d}{d t} [25])}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.8

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

$\frac{t^{2} + 10 t + 25 - (t - 5) (2 t + 10 \cdot 1 + \frac{d}{d t} [25])}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.9

Multiply $10$ by $1$ .

$\frac{t^{2} + 10 t + 25 - (t - 5) (2 t + 10 + \frac{d}{d t} [25])}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.10

Since $25$ is constant with respect to $t$ , the derivative of $25$ with respect to $t$ is $0$ .

$\frac{t^{2} + 10 t + 25 - (t - 5) (2 t + 10 + 0)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 2.11

Add $2 t + 10$ and $0$ .

$\frac{t^{2} + 10 t + 25 - (t - 5) (2 t + 10)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3

Simplify.

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Step 3.1

Apply the distributive property.

$\frac{t^{2} + 10 t + 25 + (- t - - 5) (2 t + 10)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2

Simplify the numerator.

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Step 3.2.1

Simplify each term.

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Step 3.2.1.1

Multiply $- 1$ by $- 5$ .

$\frac{t^{2} + 10 t + 25 + (- t + 5) (2 t + 10)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.2

Expand $(- t + 5) (2 t + 10)$ using the FOIL Method.

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Step 3.2.1.2.1

Apply the distributive property.

$\frac{t^{2} + 10 t + 25 - t (2 t + 10) + 5 (2 t + 10)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.2.2

Apply the distributive property.

$\frac{t^{2} + 10 t + 25 - t (2 t) - t \cdot 10 + 5 (2 t + 10)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.2.3

Apply the distributive property.

$\frac{t^{2} + 10 t + 25 - t (2 t) - t \cdot 10 + 5 (2 t) + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3

Simplify and combine like terms.

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Step 3.2.1.3.1

Simplify each term.

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Step 3.2.1.3.1.1

Rewrite using the commutative property of multiplication.

$\frac{t^{2} + 10 t + 25 - 1 \cdot 2 t \cdot t - t \cdot 10 + 5 (2 t) + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.1.2

Multiply $t$ by $t$ by adding the exponents.

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Step 3.2.1.3.1.2.1

Move $t$ .

$\frac{t^{2} + 10 t + 25 - 1 \cdot 2 (t \cdot t) - t \cdot 10 + 5 (2 t) + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.1.2.2

Multiply $t$ by $t$ .

$\frac{t^{2} + 10 t + 25 - 1 \cdot 2 t^{2} - t \cdot 10 + 5 (2 t) + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.1.3

Multiply $- 1$ by $2$ .

$\frac{t^{2} + 10 t + 25 - 2 t^{2} - t \cdot 10 + 5 (2 t) + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.1.4

Multiply $10$ by $- 1$ .

$\frac{t^{2} + 10 t + 25 - 2 t^{2} - 10 t + 5 (2 t) + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.1.5

Multiply $2$ by $5$ .

$\frac{t^{2} + 10 t + 25 - 2 t^{2} - 10 t + 10 t + 5 \cdot 10}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.1.6

Multiply $5$ by $10$ .

$\frac{t^{2} + 10 t + 25 - 2 t^{2} - 10 t + 10 t + 50}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.2

Add $- 10 t$ and $10 t$ .

$\frac{t^{2} + 10 t + 25 - 2 t^{2} + 0 + 50}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.1.3.3

Add $- 2 t^{2}$ and $0$ .

$\frac{t^{2} + 10 t + 25 - 2 t^{2} + 50}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.2

Subtract $2 t^{2}$ from $t^{2}$ .

$\frac{- t^{2} + 10 t + 25 + 50}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.2.3

Add $25$ and $50$ .

$\frac{- t^{2} + 10 t + 75}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.3

Factor by grouping.

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Step 3.3.1

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = - 1 \cdot 75 = - 75$ and whose sum is $b = 10$ .

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Step 3.3.1.1

Factor $10$ out of $10 t$ .

$\frac{- t^{2} + 10 (t) + 75}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.3.1.2

Rewrite $10$ as $- 5$ plus $15$

$\frac{- t^{2} + (- 5 + 15) t + 75}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.3.1.3

Apply the distributive property.

$\frac{- t^{2} - 5 t + 15 t + 75}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.3.2

Factor out the greatest common factor from each group.

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Step 3.3.2.1

Group the first two terms and the last two terms.

$\frac{(- t^{2} - 5 t) + 15 t + 75}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.3.2.2

Factor out the greatest common factor (GCF) from each group.

$\frac{t (- t - 5) - 15 (- t - 5)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.3.3

Factor the polynomial by factoring out the greatest common factor, $- t - 5$ .

$\frac{(- t - 5) (t - 15)}{{(t^{2} + 10 t + 25)}^{2}}$

Step 3.4

Simplify the denominator.

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Step 3.4.1

Factor using the perfect square rule.

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Step 3.4.1.1

Rewrite $25$ as $5^{2}$ .

$\frac{(- t - 5) (t - 15)}{{(t^{2} + 10 t + 5^{2})}^{2}}$

Step 3.4.1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$10 t = 2 \cdot t \cdot 5$

Step 3.4.1.3

Rewrite the polynomial.

$\frac{(- t - 5) (t - 15)}{{(t^{2} + 2 \cdot t \cdot 5 + 5^{2})}^{2}}$

Step 3.4.1.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = t$ and $b = 5$ .

$\frac{(- t - 5) (t - 15)}{{({(t + 5)}^{2})}^{2}}$

Step 3.4.2

Multiply the exponents in ${({(t + 5)}^{2})}^{2}$ .

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Step 3.4.2.1

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

$\frac{(- t - 5) (t - 15)}{{(t + 5)}^{2 \cdot 2}}$

Step 3.4.2.2

Multiply $2$ by $2$ .

$\frac{(- t - 5) (t - 15)}{{(t + 5)}^{4}}$

Step 3.4.3

Use the Binomial Theorem.

$\frac{(- t - 5) (t - 15)}{t^{4} + 4 t^{3} \cdot 5 + 6 t^{2} \cdot 5^{2} + 4 t \cdot 5^{3} + 5^{4}}$

Step 3.4.4

Simplify each term.

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Step 3.4.4.1

Multiply $5$ by $4$ .

$\frac{(- t - 5) (t - 15)}{t^{4} + 20 t^{3} + 6 t^{2} \cdot 5^{2} + 4 t \cdot 5^{3} + 5^{4}}$

Step 3.4.4.2

Raise $5$ to the power of $2$ .

$\frac{(- t - 5) (t - 15)}{t^{4} + 20 t^{3} + 6 t^{2} \cdot 25 + 4 t \cdot 5^{3} + 5^{4}}$

Step 3.4.4.3

Multiply $25$ by $6$ .

$\frac{(- t - 5) (t - 15)}{t^{4} + 20 t^{3} + 150 t^{2} + 4 t \cdot 5^{3} + 5^{4}}$

Step 3.4.4.4

Raise $5$ to the power of $3$ .

$\frac{(- t - 5) (t - 15)}{t^{4} + 20 t^{3} + 150 t^{2} + 4 t \cdot 125 + 5^{4}}$

Step 3.4.4.5

Multiply $125$ by $4$ .

$\frac{(- t - 5) (t - 15)}{t^{4} + 20 t^{3} + 150 t^{2} + 500 t + 5^{4}}$

Step 3.4.4.6

Raise $5$ to the power of $4$ .

$\frac{(- t - 5) (t - 15)}{t^{4} + 20 t^{3} + 150 t^{2} + 500 t + 625}$

Step 3.4.5

Make each term match the terms from the binomial theorem formula.

$\frac{(- t - 5) (t - 15)}{t^{4} + 4 \cdot 5 t^{3} + 6 \cdot 5^{2} t^{2} + 4 \cdot 5^{3} t + 5^{4}}$

Step 3.4.6

Factor $t^{4} + 4 \cdot 5 t^{3} + 6 \cdot 5^{2} t^{2} + 4 \cdot 5^{3} t + 5^{4}$ using the binomial theorem.

$\frac{(- t - 5) (t - 15)}{{(t + 5)}^{4}}$

Step 3.5

Cancel the common factor of $- t - 5$ and ${(t + 5)}^{4}$ .

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Step 3.5.1

Factor $- 1$ out of $- t$ .

$\frac{(- (t) - 5) (t - 15)}{{(t + 5)}^{4}}$

Step 3.5.2

Rewrite $- 5$ as $- 1 (5)$ .

$\frac{(- (t) - 1 (5)) (t - 15)}{{(t + 5)}^{4}}$

Step 3.5.3

Factor $- 1$ out of $- (t) - 1 (5)$ .

$\frac{- (t + 5) (t - 15)}{{(t + 5)}^{4}}$

Step 3.5.4

Rewrite $- (t + 5)$ as $- 1 (t + 5)$ .

$\frac{- 1 (t + 5) (t - 15)}{{(t + 5)}^{4}}$

Step 3.5.5

Factor $t + 5$ out of $- 1 (t + 5) (t - 15)$ .

$\frac{(t + 5) (- 1 (t - 15))}{{(t + 5)}^{4}}$

Step 3.5.6

Cancel the common factors.

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Step 3.5.6.1

Factor $t + 5$ out of ${(t + 5)}^{4}$ .

$\frac{(t + 5) (- 1 (t - 15))}{(t + 5) {(t + 5)}^{3}}$

Step 3.5.6.2

Cancel the common factor.

$\frac{(t + 5) (- 1 (t - 15))}{(t + 5) {(t + 5)}^{3}}$

Step 3.5.6.3

Rewrite the expression.

$\frac{- 1 (t - 15)}{{(t + 5)}^{3}}$

Step 3.6

Move the negative in front of the fraction.

$- \frac{t - 15}{{(t + 5)}^{3}}$