Calculus Examples

$x^{2} + y^{2} = 25$

Step 1

Solve the equation as $y$ in terms of $x$ .

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Step 1.1

Subtract $x^{2}$ from both sides of the equation.

$y^{2} = 25 - x^{2}$

Step 1.2

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$y = \pm \sqrt{25 - x^{2}}$

Step 1.3

Simplify $\pm \sqrt{25 - x^{2}}$ .

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Step 1.3.1

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

$y = \pm \sqrt{5^{2} - x^{2}}$

Step 1.3.2

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 $b = x$ .

$y = \pm \sqrt{(5 + x) (5 - x)}$

Step 1.4

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.4.1

First, use the positive value of the $\pm$ to find the first solution.

$y = \sqrt{(5 + x) (5 - x)}$

Step 1.4.2

Next, use the negative value of the $\pm$ to find the second solution.

$y = - \sqrt{(5 + x) (5 - x)}$

Step 1.4.3

The complete solution is the result of both the positive and negative portions of the solution.

$y = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)}$

$y = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)}$

$y = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)}$

Step 2

Set each solution of $y$ as a function of $x$ .

$y = \sqrt{(5 + x) (5 - x)} \to f (x) = \sqrt{(5 + x) (5 - x)}$

$y = - \sqrt{(5 + x) (5 - x)} \to f (x) = - \sqrt{(5 + x) (5 - x)}$

Step 3

Because the $y$ variable in the equation $x^{2} + y^{2} = 25$ has a degree greater than $1$ , use implicit differentiation to solve for the derivative $\frac{d y}{d x}$ .

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Step 3.1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} (25)$

Step 3.2

Differentiate the left side of the equation.

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Step 3.2.1

Differentiate.

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Step 3.2.1.1

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

$\frac{d}{d x} [x^{2}] + \frac{d}{d x} [y^{2}]$

Step 3.2.1.2

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

$2 x + \frac{d}{d x} [y^{2}]$

Step 3.2.2

Evaluate $\frac{d}{d x} [y^{2}]$ .

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Step 3.2.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 3.2.2.1.1

To apply the Chain Rule, set $u$ as $y$ .

$2 x + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y]$

Step 3.2.2.1.2

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

$2 x + 2 u \frac{d}{d x} [y]$

Step 3.2.2.1.3

Replace all occurrences of $u$ with $y$ .

$2 x + 2 y \frac{d}{d x} [y]$

Step 3.2.2.2

Rewrite $\frac{d}{d x} [y]$ as $y'$ .

$2 x + 2 y y'$

Step 3.2.3

Reorder terms.

$2 y y' + 2 x$

Step 3.3

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

$0$

Step 3.4

Reform the equation by setting the left side equal to the right side.

$2 y y' + 2 x = 0$

Step 3.5

Solve for $y'$ .

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Step 3.5.1

Subtract $2 x$ from both sides of the equation.

$2 y y' = - 2 x$

Step 3.5.2

Divide each term in $2 y y' = - 2 x$ by $2 y$ and simplify.

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Step 3.5.2.1

Divide each term in $2 y y' = - 2 x$ by $2 y$ .

$\frac{2 y y'}{2 y} = \frac{- 2 x}{2 y}$

Step 3.5.2.2

Simplify the left side.

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Step 3.5.2.2.1

Cancel the common factor of $2$ .

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Step 3.5.2.2.1.1

Cancel the common factor.

$\frac{2 y y'}{2 y} = \frac{- 2 x}{2 y}$

Step 3.5.2.2.1.2

Rewrite the expression.

$\frac{y y'}{y} = \frac{- 2 x}{2 y}$

Step 3.5.2.2.2

Cancel the common factor of $y$ .

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Step 3.5.2.2.2.1

Cancel the common factor.

$\frac{y y'}{y} = \frac{- 2 x}{2 y}$

Step 3.5.2.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{- 2 x}{2 y}$

Step 3.5.2.3

Simplify the right side.

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Step 3.5.2.3.1

Cancel the common factor of $- 2$ and $2$ .

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Step 3.5.2.3.1.1

Factor $2$ out of $- 2 x$ .

$y' = \frac{2 (- x)}{2 y}$

Step 3.5.2.3.1.2

Cancel the common factors.

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Step 3.5.2.3.1.2.1

Factor $2$ out of $2 y$ .

$y' = \frac{2 (- x)}{2 (y)}$

Step 3.5.2.3.1.2.2

Cancel the common factor.

$y' = \frac{2 (- x)}{2 y}$

Step 3.5.2.3.1.2.3

Rewrite the expression.

$y' = \frac{- x}{y}$

Step 3.5.2.3.2

Move the negative in front of the fraction.

$y' = - \frac{x}{y}$

Step 3.6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = - \frac{x}{y}$

Step 4

Set the numerator equal to zero.

$x = 0$

Step 5

Solve the function $f (x) = \sqrt{(5 + x) (5 - x)}$ at $x = 0$ .

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Step 5.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = \sqrt{(5 + 0) (5 - (0))}$

Step 5.2

Simplify the result.

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Step 5.2.1

Remove parentheses.

$f (0) = \sqrt{(5 + 0) (5 - (0))}$

Step 5.2.2

Add $5$ and $0$ .

$f (0) = \sqrt{5 (5 - (0))}$

Step 5.2.3

Multiply $- 1$ by $0$ .

$f (0) = \sqrt{5 (5 + 0)}$

Step 5.2.4

Add $5$ and $0$ .

$f (0) = \sqrt{5 \cdot 5}$

Step 5.2.5

Multiply $5$ by $5$ .

$f (0) = \sqrt{25}$

Step 5.2.6

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

$f (0) = \sqrt{5^{2}}$

Step 5.2.7

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = 5$

Step 5.2.8

The final answer is $5$ .

$5$

Step 6

Solve the function $f (x) = - \sqrt{(5 + x) (5 - x)}$ at $x = 0$ .

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Step 6.1

Replace the variable $x$ with $0$ in the expression.

$f (0) = - \sqrt{(5 + 0) (5 - (0))}$

Step 6.2

Simplify the result.

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Step 6.2.1

Remove parentheses.

$f (0) = - \sqrt{(5 + 0) (5 - (0))}$

Step 6.2.2

Add $5$ and $0$ .

$f (0) = - \sqrt{5 (5 - (0))}$

Step 6.2.3

Multiply $- 1$ by $0$ .

$f (0) = - \sqrt{5 (5 + 0)}$

Step 6.2.4

Add $5$ and $0$ .

$f (0) = - \sqrt{5 \cdot 5}$

Step 6.2.5

Multiply $5$ by $5$ .

$f (0) = - \sqrt{25}$

Step 6.2.6

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

$f (0) = - \sqrt{5^{2}}$

Step 6.2.7

Pull terms out from under the radical, assuming positive real numbers.

$f (0) = - 1 \cdot 5$

Step 6.2.8

Multiply $- 1$ by $5$ .

$f (0) = - 5$

Step 6.2.9

The final answer is $- 5$ .

$- 5$

Step 7

The horizontal tangent lines are $y = 5, y = - 5$

$y = 5, y = - 5$

Step 8