Calculus Examples

$y = \frac{x^{2} + 5 x}{25 - x^{2}}$

Step 1

Write $y = \frac{x^{2} + 5 x}{25 - x^{2}}$ as a function.

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

Step 2

Find the first derivative.

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Step 2.1

Find the first derivative.

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Step 2.1.1

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

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

Step 2.1.2

Differentiate.

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Step 2.1.2.1

By the Sum Rule, the derivative of $x^{2} + 5 x$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [5 x]$ .

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

Step 2.1.2.2

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

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

Step 2.1.2.3

Since $5$ is constant with respect to $x$ , the derivative of $5 x$ with respect to $x$ is $5 \frac{d}{d x} [x]$ .

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

Step 2.1.2.4

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

$f' (x) = \frac{(25 - x^{2}) (2 x + 5 \cdot 1) - (x^{2} + 5 x) \frac{d}{d x} (25 - x^{2})}{{(25 - x^{2})}^{2}}$

Step 2.1.2.5

Multiply $5$ by $1$ .

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

Step 2.1.2.6

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

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

Step 2.1.2.7

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

$f' (x) = \frac{(25 - x^{2}) (2 x + 5) - (x^{2} + 5 x) (0 + \frac{d}{d x} (- x^{2}))}{{(25 - x^{2})}^{2}}$

Step 2.1.2.8

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

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

Step 2.1.2.9

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

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

Step 2.1.2.10

Multiply.

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Step 2.1.2.10.1

Multiply $- 1$ by $- 1$ .

$f' (x) = \frac{(25 - x^{2}) (2 x + 5) + 1 (x^{2} + 5 x) \frac{d}{d x} (x^{2})}{{(25 - x^{2})}^{2}}$

Step 2.1.2.10.2

Multiply $x^{2} + 5 x$ by $1$ .

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

Step 2.1.2.11

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

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

Step 2.1.2.12

Move $2$ to the left of $x^{2} + 5 x$ .

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

Step 2.1.3

Simplify.

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Step 2.1.3.1

Apply the distributive property.

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

Step 2.1.3.2

Apply the distributive property.

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

Step 2.1.3.3

Simplify the numerator.

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Step 2.1.3.3.1

Simplify each term.

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Step 2.1.3.3.1.1

Expand $(25 - x^{2}) (2 x + 5)$ using the FOIL Method.

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Step 2.1.3.3.1.1.1

Apply the distributive property.

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

Step 2.1.3.3.1.1.2

Apply the distributive property.

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

Step 2.1.3.3.1.1.3

Apply the distributive property.

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

Step 2.1.3.3.1.2

Simplify each term.

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Step 2.1.3.3.1.2.1

Multiply $2$ by $25$ .

$f' (x) = \frac{50 x + 25 \cdot 5 - x^{2} (2 x) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.2

Multiply $25$ by $5$ .

$f' (x) = \frac{50 x + 125 - x^{2} (2 x) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.3

Rewrite using the commutative property of multiplication.

$f' (x) = \frac{50 x + 125 - 1 \cdot (2 x^{2} x) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.4

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.1.3.3.1.2.4.1

Move $x$ .

$f' (x) = \frac{50 x + 125 - 1 \cdot (2 (x \cdot x^{2})) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.4.2

Multiply $x$ by $x^{2}$ .

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Step 2.1.3.3.1.2.4.2.1

Raise $x$ to the power of $1$ .

$f' (x) = \frac{50 x + 125 - 1 \cdot (2 (x \cdot x^{2})) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.4.2.2

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

$f' (x) = \frac{50 x + 125 - 1 \cdot (2 x^{1 + 2}) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.4.3

Add $1$ and $2$ .

$f' (x) = \frac{50 x + 125 - 1 \cdot (2 x^{3}) - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.5

Multiply $- 1$ by $2$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - x^{2} \cdot 5 + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.2.6

Multiply $5$ by $- 1$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{2} x + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.3

Multiply $x^{2}$ by $x$ by adding the exponents.

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Step 2.1.3.3.1.3.1

Move $x$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 (x \cdot x^{2}) + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.3.2

Multiply $x$ by $x^{2}$ .

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Step 2.1.3.3.1.3.2.1

Raise $x$ to the power of $1$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 (x \cdot x^{2}) + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.3.2.2

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

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{1 + 2} + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.3.3

Add $1$ and $2$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{3} + 2 (5 x) x}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.4

Multiply $x$ by $x$ by adding the exponents.

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Step 2.1.3.3.1.4.1

Move $x$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{3} + 2 (5 (x \cdot x))}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.4.2

Multiply $x$ by $x$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{3} + 2 (5 x^{2})}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.1.5

Multiply $5$ by $2$ .

$f' (x) = \frac{50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{3} + 10 x^{2}}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.2

Combine the opposite terms in $50 x + 125 - 2 x^{3} - 5 x^{2} + 2 x^{3} + 10 x^{2}$ .

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Step 2.1.3.3.2.1

Add $- 2 x^{3}$ and $2 x^{3}$ .

$f' (x) = \frac{50 x + 125 - 5 x^{2} + 0 + 10 x^{2}}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.2.2

Add $50 x + 125 - 5 x^{2}$ and $0$ .

$f' (x) = \frac{50 x + 125 - 5 x^{2} + 10 x^{2}}{{(25 - x^{2})}^{2}}$

Step 2.1.3.3.3

Add $- 5 x^{2}$ and $10 x^{2}$ .

$f' (x) = \frac{50 x + 125 + 5 x^{2}}{{(25 - x^{2})}^{2}}$

Step 2.1.3.4

Reorder terms.

$f' (x) = \frac{5 x^{2} + 50 x + 125}{{(- x^{2} + 25)}^{2}}$

Step 2.1.3.5

Simplify the numerator.

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Step 2.1.3.5.1

Factor $5$ out of $5 x^{2} + 50 x + 125$ .

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Step 2.1.3.5.1.1

Factor $5$ out of $5 x^{2}$ .

$f' (x) = \frac{5 (x^{2}) + 50 x + 125}{{(- x^{2} + 25)}^{2}}$

Step 2.1.3.5.1.2

Factor $5$ out of $50 x$ .

$f' (x) = \frac{5 (x^{2}) + 5 (10 x) + 125}{{(- x^{2} + 25)}^{2}}$

Step 2.1.3.5.1.3

Factor $5$ out of $125$ .

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

Step 2.1.3.5.1.4

Factor $5$ out of $5 x^{2} + 5 (10 x)$ .

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

Step 2.1.3.5.1.5

Factor $5$ out of $5 (x^{2} + 10 x) + 5 \cdot 25$ .

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

Step 2.1.3.5.2

Factor using the perfect square rule.

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Step 2.1.3.5.2.1

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

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

Step 2.1.3.5.2.2

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

$10 x = 2 \cdot x \cdot 5$

Step 2.1.3.5.2.3

Rewrite the polynomial.

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

Step 2.1.3.5.2.4

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

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

Step 2.1.3.6

Simplify the denominator.

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Step 2.1.3.6.1

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

$f' (x) = \frac{5 {(x + 5)}^{2}}{{(- x^{2} + 5^{2})}^{2}}$

Step 2.1.3.6.2

Reorder $- x^{2}$ and $5^{2}$ .

$f' (x) = \frac{5 {(x + 5)}^{2}}{{(5^{2} - x^{2})}^{2}}$

Step 2.1.3.6.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 5$ and $b = x$ .

$f' (x) = \frac{5 {(x + 5)}^{2}}{{((5 + x) (5 - x))}^{2}}$

Step 2.1.3.6.4

Apply the product rule to $(5 + x) (5 - x)$ .

$f' (x) = \frac{5 {(x + 5)}^{2}}{{(5 + x)}^{2} {(5 - x)}^{2}}$

Step 2.1.3.7

Cancel the common factor of ${(x + 5)}^{2}$ and ${(5 + x)}^{2}$ .

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Step 2.1.3.7.1

Reorder terms.

$f' (x) = \frac{5 {(x + 5)}^{2}}{{(x + 5)}^{2} {(5 - x)}^{2}}$

Step 2.1.3.7.2

Cancel the common factor.

$f' (x) = \frac{5 {(x + 5)}^{2}}{{(x + 5)}^{2} {(5 - x)}^{2}}$

Step 2.1.3.7.3

Rewrite the expression.

$f' (x) = \frac{5}{{(5 - x)}^{2}}$

Step 2.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{5}{{(5 - x)}^{2}}$ .

$\frac{5}{{(5 - x)}^{2}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $\frac{5}{{(5 - x)}^{2}} = 0$ .

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Step 3.1

Set the first derivative equal to $0$ .

$\frac{5}{{(5 - x)}^{2}} = 0$

Step 3.2

Set the numerator equal to zero.

$5 = 0$

Step 3.3

Since $5 \neq 0$ , there are no solutions.

No solution

Step 4

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 5

Find where the derivative is undefined.

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Step 5.1

Set the denominator in $\frac{5}{{(5 - x)}^{2}}$ equal to $0$ to find where the expression is undefined.

${(5 - x)}^{2} = 0$

Step 5.2

Solve for $x$ .

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Step 5.2.1

Set the $5 - x$ equal to $0$ .

$5 - x = 0$

Step 5.2.2

Solve for $x$ .

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Step 5.2.2.1

Subtract $5$ from both sides of the equation.

$- x = - 5$

Step 5.2.2.2

Divide each term in $- x = - 5$ by $- 1$ and simplify.

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Step 5.2.2.2.1

Divide each term in $- x = - 5$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 5}{- 1}$

Step 5.2.2.2.2

Simplify the left side.

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Step 5.2.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 5}{- 1}$

Step 5.2.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 5}{- 1}$

Step 5.2.2.2.3

Simplify the right side.

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Step 5.2.2.2.3.1

Divide $- 5$ by $- 1$ .

$x = 5$

Step 6

After finding the point that makes the derivative $f' (x) = \frac{5}{{(5 - x)}^{2}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{x^{2} + 5 x}{25 - x^{2}}$ is increasing and where it is decreasing is $(- \infty, 5) \cup (5, \infty)$ .

$(- \infty, 5) \cup (5, \infty)$

Step 7

Substitute a value from the interval $(- \infty, 5)$ into the derivative to determine if the function is increasing or decreasing.

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Step 7.1

Replace the variable $x$ with $4$ in the expression.

$f' (4) = \frac{5}{{(5 - (4))}^{2}}$

Step 7.2

Simplify the result.

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Step 7.2.1

Simplify the denominator.

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Step 7.2.1.1

Multiply $- 1$ by $4$ .

$f' (4) = \frac{5}{{(5 - 4)}^{2}}$

Step 7.2.1.2

Subtract $4$ from $5$ .

$f' (4) = \frac{5}{1^{2}}$

Step 7.2.1.3

One to any power is one.

$f' (4) = \frac{5}{1}$

Step 7.2.2

Divide $5$ by $1$ .

$f' (4) = 5$

Step 7.2.3

The final answer is $5$ .

$5$

Step 7.3

At $x = 4$ the derivative is $5$ . Since this is positive, the function is increasing on $(- \infty, 5)$ .

Increasing on $(- \infty, 5)$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(5, \infty)$ into the derivative to determine if the function is increasing or decreasing.

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Step 8.1

Replace the variable $x$ with $6$ in the expression.

$f' (6) = \frac{5}{{(5 - (6))}^{2}}$

Step 8.2

Simplify the result.

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Step 8.2.1

Simplify the denominator.

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Step 8.2.1.1

Multiply $- 1$ by $6$ .

$f' (6) = \frac{5}{{(5 - 6)}^{2}}$

Step 8.2.1.2

Subtract $6$ from $5$ .

$f' (6) = \frac{5}{{(- 1)}^{2}}$

Step 8.2.1.3

Raise $- 1$ to the power of $2$ .

$f' (6) = \frac{5}{1}$

Step 8.2.2

Divide $5$ by $1$ .

$f' (6) = 5$

Step 8.2.3

The final answer is $5$ .

$5$

Step 8.3

At $x = 6$ the derivative is $5$ . Since this is positive, the function is increasing on $(5, \infty)$ .

Increasing on $(5, \infty)$ since $f' (x) > 0$

Step 9

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, 5), (5, \infty)$

Step 10