Calculus Examples

$y = \frac{2 - v^{2}}{2 + v^{2}}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y'} y = \frac{d}{d y'} \cdot \frac{2 - {y'}^{2}}{2 + {y'}^{2}}$

Step 2

The derivative of $y$ with respect to $y'$ is $y'$ .

$y'$

Step 3

Differentiate the right side of the equation.

Tap for more steps...

Step 3.1

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

$\frac{(2 + {y'}^{2}) \frac{d}{d v} [2 - {y'}^{2}] - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2

Differentiate.

Tap for more steps...

Step 3.2.1

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

$\frac{(2 + {y'}^{2}) (\frac{d}{d v} [2] + \frac{d}{d v} [- {y'}^{2}]) - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.2

Since $2$ is constant with respect to $y'$ , the derivative of $2$ with respect to $y'$ is $0$ .

$\frac{(2 + {y'}^{2}) (0 + \frac{d}{d v} [- {y'}^{2}]) - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.3

Add $0$ and $\frac{d}{d v} [- {y'}^{2}]$ .

$\frac{(2 + {y'}^{2}) \frac{d}{d v} [- {y'}^{2}] - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.4

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

$\frac{(2 + {y'}^{2}) (- \frac{d}{d v} [{y'}^{2}]) - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.5

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

$\frac{(2 + {y'}^{2}) (- (2 y')) - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.6

Multiply $2$ by $- 1$ .

$\frac{(2 + {y'}^{2}) (- 2 y') - (2 - {y'}^{2}) \frac{d}{d v} [2 + {y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.7

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

$\frac{(2 + {y'}^{2}) (- 2 y') - (2 - {y'}^{2}) (\frac{d}{d v} [2] + \frac{d}{d v} [{y'}^{2}])}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.8

Since $2$ is constant with respect to $y'$ , the derivative of $2$ with respect to $y'$ is $0$ .

$\frac{(2 + {y'}^{2}) (- 2 y') - (2 - {y'}^{2}) (0 + \frac{d}{d v} [{y'}^{2}])}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.9

Add $0$ and $\frac{d}{d v} [{y'}^{2}]$ .

$\frac{(2 + {y'}^{2}) (- 2 y') - (2 - {y'}^{2}) \frac{d}{d v} [{y'}^{2}]}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.10

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

$\frac{(2 + {y'}^{2}) (- 2 y') - (2 - {y'}^{2}) (2 y')}{{(2 + {y'}^{2})}^{2}}$

Step 3.2.11

Multiply $2$ by $- 1$ .

$\frac{(2 + {y'}^{2}) (- 2 y') - 2 (2 - {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3

Simplify.

Tap for more steps...

Step 3.3.1

Apply the distributive property.

$\frac{2 (- 2 y') + {y'}^{2} (- 2 y') - 2 (2 - {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.2

Apply the distributive property.

$\frac{2 (- 2 y') + {y'}^{2} (- 2 y') + (- 2 \cdot 2 - 2 (- {y'}^{2})) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.3

Apply the distributive property.

$\frac{2 (- 2 y') + {y'}^{2} (- 2 y') - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4

Simplify the numerator.

Tap for more steps...

Step 3.3.4.1

Simplify each term.

Tap for more steps...

Step 3.3.4.1.1

Multiply $- 2$ by $2$ .

$\frac{- 4 y' + {y'}^{2} (- 2 y') - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.2

Rewrite using the commutative property of multiplication.

$\frac{- 4 y' - 2 {y'}^{2} y' - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.3

Multiply ${y'}^{2}$ by $y'$ by adding the exponents.

Tap for more steps...

Step 3.3.4.1.3.1

Move $y'$ .

$\frac{- 4 y' - 2 (y' {y'}^{2}) - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.3.2

Multiply $y'$ by ${y'}^{2}$ .

Tap for more steps...

Step 3.3.4.1.3.2.1

Raise $y'$ to the power of $1$ .

$\frac{- 4 y' - 2 ({y'}^{1} {y'}^{2}) - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.3.2.2

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

$\frac{- 4 y' - 2 {y'}^{1 + 2} - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.3.3

Add $1$ and $2$ .

$\frac{- 4 y' - 2 {y'}^{3} - 2 \cdot 2 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.4

Multiply $- 2$ by $2$ .

$\frac{- 4 y' - 2 {y'}^{3} - 4 y' - 2 (- {y'}^{2}) y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.5

Multiply ${y'}^{2}$ by $y'$ by adding the exponents.

Tap for more steps...

Step 3.3.4.1.5.1

Move $y'$ .

$\frac{- 4 y' - 2 {y'}^{3} - 4 y' - 2 (- (y' {y'}^{2}))}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.5.2

Multiply $y'$ by ${y'}^{2}$ .

Tap for more steps...

Step 3.3.4.1.5.2.1

Raise $y'$ to the power of $1$ .

$\frac{- 4 y' - 2 {y'}^{3} - 4 y' - 2 (- ({y'}^{1} {y'}^{2}))}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.5.2.2

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

$\frac{- 4 y' - 2 {y'}^{3} - 4 y' - 2 (- {y'}^{1 + 2})}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.5.3

Add $1$ and $2$ .

$\frac{- 4 y' - 2 {y'}^{3} - 4 y' - 2 (- {y'}^{3})}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.1.6

Multiply $- 1$ by $- 2$ .

$\frac{- 4 y' - 2 {y'}^{3} - 4 y' + 2 {y'}^{3}}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.2

Combine the opposite terms in $- 4 y' - 2 {y'}^{3} - 4 y' + 2 {y'}^{3}$ .

Tap for more steps...

Step 3.3.4.2.1

Add $- 2 {y'}^{3}$ and $2 {y'}^{3}$ .

$\frac{- 4 y' - 4 y' + 0}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.2.2

Add $- 4 y' - 4 y'$ and $0$ .

$\frac{- 4 y' - 4 y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.4.3

Subtract $4 y'$ from $- 4 y'$ .

$\frac{- 8 y'}{{(2 + {y'}^{2})}^{2}}$

Step 3.3.5

Move the negative in front of the fraction.

$- \frac{8 y'}{{(2 + {y'}^{2})}^{2}}$

Step 4

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

$y' = - \frac{8 y'}{{(2 + {y'}^{2})}^{2}}$

Step 5

Solve for $y'$ .

Tap for more steps...

Step 5.1

Find the LCD of the terms in the equation.

Tap for more steps...

Step 5.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, {(2 + {y'}^{2})}^{2}$

Step 5.1.2

The LCM of one and any expression is the expression.

${(2 + {y'}^{2})}^{2}$

Step 5.2

Multiply each term in $y' = - \frac{8 y'}{{(2 + {y'}^{2})}^{2}}$ by ${(2 + {y'}^{2})}^{2}$ to eliminate the fractions.

Tap for more steps...

Step 5.2.1

Multiply each term in $y' = - \frac{8 y'}{{(2 + {y'}^{2})}^{2}}$ by ${(2 + {y'}^{2})}^{2}$ .

$y' {(2 + {y'}^{2})}^{2} = - \frac{8 y'}{{(2 + {y'}^{2})}^{2}} {(2 + {y'}^{2})}^{2}$

Step 5.2.2

Simplify the right side.

Tap for more steps...

Step 5.2.2.1

Cancel the common factor of ${(2 + {y'}^{2})}^{2}$ .

Tap for more steps...

Step 5.2.2.1.1

Move the leading negative in $- \frac{8 y'}{{(2 + {y'}^{2})}^{2}}$ into the numerator.

$y' {(2 + {y'}^{2})}^{2} = \frac{- 8 y'}{{(2 + {y'}^{2})}^{2}} {(2 + {y'}^{2})}^{2}$

Step 5.2.2.1.2

Cancel the common factor.

$y' {(2 + {y'}^{2})}^{2} = \frac{- 8 y'}{{(2 + {y'}^{2})}^{2}} {(2 + {y'}^{2})}^{2}$

Step 5.2.2.1.3

Rewrite the expression.

$y' {(2 + {y'}^{2})}^{2} = - 8 y'$

Step 5.3

Solve the equation.

Tap for more steps...

Step 5.3.1

Move all terms containing $y'$ to the left side of the equation.

Tap for more steps...

Step 5.3.1.1

Add $8 y'$ to both sides of the equation.

$y' {(2 + {y'}^{2})}^{2} + 8 y' = 0$

Step 5.3.1.2

Simplify each term.

Tap for more steps...

Step 5.3.1.2.1

Rewrite ${(2 + {y'}^{2})}^{2}$ as $(2 + {y'}^{2}) (2 + {y'}^{2})$ .

$y' ((2 + {y'}^{2}) (2 + {y'}^{2})) + 8 y' = 0$

Step 5.3.1.2.2

Expand $(2 + {y'}^{2}) (2 + {y'}^{2})$ using the FOIL Method.

Tap for more steps...

Step 5.3.1.2.2.1

Apply the distributive property.

$y' (2 (2 + {y'}^{2}) + {y'}^{2} (2 + {y'}^{2})) + 8 y' = 0$

Step 5.3.1.2.2.2

Apply the distributive property.

$y' (2 \cdot 2 + 2 {y'}^{2} + {y'}^{2} (2 + {y'}^{2})) + 8 y' = 0$

Step 5.3.1.2.2.3

Apply the distributive property.

$y' (2 \cdot 2 + 2 {y'}^{2} + {y'}^{2} \cdot 2 + {y'}^{2} {y'}^{2}) + 8 y' = 0$

Step 5.3.1.2.3

Simplify and combine like terms.

Tap for more steps...

Step 5.3.1.2.3.1

Simplify each term.

Tap for more steps...

Step 5.3.1.2.3.1.1

Multiply $2$ by $2$ .

$y' (4 + 2 {y'}^{2} + {y'}^{2} \cdot 2 + {y'}^{2} {y'}^{2}) + 8 y' = 0$

Step 5.3.1.2.3.1.2

Move $2$ to the left of ${y'}^{2}$ .

$y' (4 + 2 {y'}^{2} + 2 \cdot {y'}^{2} + {y'}^{2} {y'}^{2}) + 8 y' = 0$

Step 5.3.1.2.3.1.3

Multiply ${y'}^{2}$ by ${y'}^{2}$ by adding the exponents.

Tap for more steps...

Step 5.3.1.2.3.1.3.1

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

$y' (4 + 2 {y'}^{2} + 2 {y'}^{2} + {y'}^{2 + 2}) + 8 y' = 0$

Step 5.3.1.2.3.1.3.2

Add $2$ and $2$ .

$y' (4 + 2 {y'}^{2} + 2 {y'}^{2} + {y'}^{4}) + 8 y' = 0$

Step 5.3.1.2.3.2

Add $2 {y'}^{2}$ and $2 {y'}^{2}$ .

$y' (4 + 4 {y'}^{2} + {y'}^{4}) + 8 y' = 0$

Step 5.3.1.2.4

Apply the distributive property.

$y' \cdot 4 + y' (4 {y'}^{2}) + y' ({y'}^{4}) + 8 y' = 0$

Step 5.3.1.2.5

Simplify.

Tap for more steps...

Step 5.3.1.2.5.1

Move $4$ to the left of $y'$ .

$4 \cdot y' + y' (4 {y'}^{2}) + y' ({y'}^{4}) + 8 y' = 0$

Step 5.3.1.2.5.2

Rewrite using the commutative property of multiplication.

$4 \cdot y' + 4 y' {y'}^{2} + y' {y'}^{4} + 8 y' = 0$

Step 5.3.1.2.5.3

Multiply $y'$ by ${y'}^{4}$ by adding the exponents.

Tap for more steps...

Step 5.3.1.2.5.3.1

Multiply $y'$ by ${y'}^{4}$ .

Tap for more steps...

Step 5.3.1.2.5.3.1.1

Raise $y'$ to the power of $1$ .

$4 \cdot y' + 4 y' {y'}^{2} + {y'}^{1} {y'}^{4} + 8 y' = 0$

Step 5.3.1.2.5.3.1.2

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

$4 \cdot y' + 4 y' {y'}^{2} + {y'}^{1 + 4} + 8 y' = 0$

Step 5.3.1.2.5.3.2

Add $1$ and $4$ .

$4 \cdot y' + 4 y' {y'}^{2} + {y'}^{5} + 8 y' = 0$

Step 5.3.1.2.6

Multiply $y'$ by ${y'}^{2}$ by adding the exponents.

Tap for more steps...

Step 5.3.1.2.6.1

Move ${y'}^{2}$ .

$4 y' + 4 ({y'}^{2} y') + {y'}^{5} + 8 y' = 0$

Step 5.3.1.2.6.2

Multiply ${y'}^{2}$ by $y'$ .

Tap for more steps...

Step 5.3.1.2.6.2.1

Raise $y'$ to the power of $1$ .

$4 y' + 4 ({y'}^{2} {y'}^{1}) + {y'}^{5} + 8 y' = 0$

Step 5.3.1.2.6.2.2

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

$4 y' + 4 {y'}^{2 + 1} + {y'}^{5} + 8 y' = 0$

Step 5.3.1.2.6.3

Add $2$ and $1$ .

$4 y' + 4 {y'}^{3} + {y'}^{5} + 8 y' = 0$

Step 5.3.1.3

Add $4 y'$ and $8 y'$ .

$4 {y'}^{3} + {y'}^{5} + 12 y' = 0$

Step 5.3.2

Factor the left side of the equation.

Tap for more steps...

Step 5.3.2.1

Factor $y'$ out of $4 {y'}^{3} + {y'}^{5} + 12 y'$ .

Tap for more steps...

Step 5.3.2.1.1

Factor $y'$ out of $4 {y'}^{3}$ .

$y' (4 {y'}^{2}) + {y'}^{5} + 12 y' = 0$

Step 5.3.2.1.2

Factor $y'$ out of ${y'}^{5}$ .

$y' (4 {y'}^{2}) + y' ({y'}^{4}) + 12 y' = 0$

Step 5.3.2.1.3

Factor $y'$ out of $12 y'$ .

$y' (4 {y'}^{2}) + y' ({y'}^{4}) + y' \cdot 12 = 0$

Step 5.3.2.1.4

Factor $y'$ out of $y' (4 {y'}^{2}) + y' ({y'}^{4})$ .

$y' (4 {y'}^{2} + {y'}^{4}) + y' \cdot 12 = 0$

Step 5.3.2.1.5

Factor $y'$ out of $y' (4 {y'}^{2} + {y'}^{4}) + y' \cdot 12$ .

$y' (4 {y'}^{2} + {y'}^{4} + 12) = 0$

Step 5.3.2.2

Reorder terms.

$y' ({y'}^{4} + 4 {y'}^{2} + 12) = 0$

Step 5.3.3

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$y' = 0$

${y'}^{4} + 4 {y'}^{2} + 12 = 0$

Step 5.3.4

Set $y'$ equal to $0$ .

$y' = 0$

Step 5.3.5

Set ${y'}^{4} + 4 {y'}^{2} + 12$ equal to $0$ and solve for $y'$ .

Tap for more steps...

Step 5.3.5.1

Set ${y'}^{4} + 4 {y'}^{2} + 12$ equal to $0$ .

${y'}^{4} + 4 {y'}^{2} + 12 = 0$

Step 5.3.5.2

Solve ${y'}^{4} + 4 {y'}^{2} + 12 = 0$ for $y'$ .

Tap for more steps...

Step 5.3.5.2.1

Substitute $u = {y'}^{2}$ into the equation. This will make the quadratic formula easy to use.

$u^{2} + 4 u + 12 = 0$

$u = {y'}^{2}$

Step 5.3.5.2.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5.3.5.2.3

Substitute the values $a = 1$ , $b = 4$ , and $c = 12$ into the quadratic formula and solve for $u$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (1 \cdot 12)}}{2 \cdot 1}$

Step 5.3.5.2.4

Simplify.

Tap for more steps...

Step 5.3.5.2.4.1

Simplify the numerator.

Tap for more steps...

Step 5.3.5.2.4.1.1

Raise $4$ to the power of $2$ .

$u = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Step 5.3.5.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 12$ .

Tap for more steps...

Step 5.3.5.2.4.1.2.1

Multiply $- 4$ by $1$ .

$u = \frac{- 4 \pm \sqrt{16 - 4 \cdot 12}}{2 \cdot 1}$

Step 5.3.5.2.4.1.2.2

Multiply $- 4$ by $12$ .

$u = \frac{- 4 \pm \sqrt{16 - 48}}{2 \cdot 1}$

Step 5.3.5.2.4.1.3

Subtract $48$ from $16$ .

$u = \frac{- 4 \pm \sqrt{- 32}}{2 \cdot 1}$

Step 5.3.5.2.4.1.4

Rewrite $- 32$ as $- 1 (32)$ .

$u = \frac{- 4 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 1}$

Step 5.3.5.2.4.1.5

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$u = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 1}$

Step 5.3.5.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$u = \frac{- 4 \pm i \cdot \sqrt{32}}{2 \cdot 1}$

Step 5.3.5.2.4.1.7

Rewrite $32$ as $4^{2} \cdot 2$ .

Tap for more steps...

Step 5.3.5.2.4.1.7.1

Factor $16$ out of $32$ .

$u = \frac{- 4 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 1}$

Step 5.3.5.2.4.1.7.2

Rewrite $16$ as $4^{2}$ .

$u = \frac{- 4 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Step 5.3.5.2.4.1.8

Pull terms out from under the radical.

$u = \frac{- 4 \pm i \cdot (4 \sqrt{2})}{2 \cdot 1}$

Step 5.3.5.2.4.1.9

Move $4$ to the left of $i$ .

$u = \frac{- 4 \pm 4 i \sqrt{2}}{2 \cdot 1}$

Step 5.3.5.2.4.2

Multiply $2$ by $1$ .

$u = \frac{- 4 \pm 4 i \sqrt{2}}{2}$

Step 5.3.5.2.4.3

Simplify $\frac{- 4 \pm 4 i \sqrt{2}}{2}$ .

$u = - 2 \pm 2 i \sqrt{2}$

Step 5.3.5.2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Tap for more steps...

Step 5.3.5.2.5.1

Simplify the numerator.

Tap for more steps...

Step 5.3.5.2.5.1.1

Raise $4$ to the power of $2$ .

$u = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Step 5.3.5.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 12$ .

Tap for more steps...

Step 5.3.5.2.5.1.2.1

Multiply $- 4$ by $1$ .

$u = \frac{- 4 \pm \sqrt{16 - 4 \cdot 12}}{2 \cdot 1}$

Step 5.3.5.2.5.1.2.2

Multiply $- 4$ by $12$ .

$u = \frac{- 4 \pm \sqrt{16 - 48}}{2 \cdot 1}$

Step 5.3.5.2.5.1.3

Subtract $48$ from $16$ .

$u = \frac{- 4 \pm \sqrt{- 32}}{2 \cdot 1}$

Step 5.3.5.2.5.1.4

Rewrite $- 32$ as $- 1 (32)$ .

$u = \frac{- 4 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 1}$

Step 5.3.5.2.5.1.5

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$u = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 1}$

Step 5.3.5.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$u = \frac{- 4 \pm i \cdot \sqrt{32}}{2 \cdot 1}$

Step 5.3.5.2.5.1.7

Rewrite $32$ as $4^{2} \cdot 2$ .

Tap for more steps...

Step 5.3.5.2.5.1.7.1

Factor $16$ out of $32$ .

$u = \frac{- 4 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 1}$

Step 5.3.5.2.5.1.7.2

Rewrite $16$ as $4^{2}$ .

$u = \frac{- 4 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Step 5.3.5.2.5.1.8

Pull terms out from under the radical.

$u = \frac{- 4 \pm i \cdot (4 \sqrt{2})}{2 \cdot 1}$

Step 5.3.5.2.5.1.9

Move $4$ to the left of $i$ .

$u = \frac{- 4 \pm 4 i \sqrt{2}}{2 \cdot 1}$

Step 5.3.5.2.5.2

Multiply $2$ by $1$ .

$u = \frac{- 4 \pm 4 i \sqrt{2}}{2}$

Step 5.3.5.2.5.3

Simplify $\frac{- 4 \pm 4 i \sqrt{2}}{2}$ .

$u = - 2 \pm 2 i \sqrt{2}$

Step 5.3.5.2.5.4

Change the $\pm$ to $+$ .

$u = - 2 + 2 i \sqrt{2}$

Step 5.3.5.2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Tap for more steps...

Step 5.3.5.2.6.1

Simplify the numerator.

Tap for more steps...

Step 5.3.5.2.6.1.1

Raise $4$ to the power of $2$ .

$u = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$

Step 5.3.5.2.6.1.2

Multiply $- 4 \cdot 1 \cdot 12$ .

Tap for more steps...

Step 5.3.5.2.6.1.2.1

Multiply $- 4$ by $1$ .

$u = \frac{- 4 \pm \sqrt{16 - 4 \cdot 12}}{2 \cdot 1}$

Step 5.3.5.2.6.1.2.2

Multiply $- 4$ by $12$ .

$u = \frac{- 4 \pm \sqrt{16 - 48}}{2 \cdot 1}$

Step 5.3.5.2.6.1.3

Subtract $48$ from $16$ .

$u = \frac{- 4 \pm \sqrt{- 32}}{2 \cdot 1}$

Step 5.3.5.2.6.1.4

Rewrite $- 32$ as $- 1 (32)$ .

$u = \frac{- 4 \pm \sqrt{- 1 \cdot 32}}{2 \cdot 1}$

Step 5.3.5.2.6.1.5

Rewrite $\sqrt{- 1 (32)}$ as $\sqrt{- 1} \cdot \sqrt{32}$ .

$u = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{32}}{2 \cdot 1}$

Step 5.3.5.2.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$u = \frac{- 4 \pm i \cdot \sqrt{32}}{2 \cdot 1}$

Step 5.3.5.2.6.1.7

Rewrite $32$ as $4^{2} \cdot 2$ .

Tap for more steps...

Step 5.3.5.2.6.1.7.1

Factor $16$ out of $32$ .

$u = \frac{- 4 \pm i \cdot \sqrt{16 (2)}}{2 \cdot 1}$

Step 5.3.5.2.6.1.7.2

Rewrite $16$ as $4^{2}$ .

$u = \frac{- 4 \pm i \cdot \sqrt{4^{2} \cdot 2}}{2 \cdot 1}$

Step 5.3.5.2.6.1.8

Pull terms out from under the radical.

$u = \frac{- 4 \pm i \cdot (4 \sqrt{2})}{2 \cdot 1}$

Step 5.3.5.2.6.1.9

Move $4$ to the left of $i$ .

$u = \frac{- 4 \pm 4 i \sqrt{2}}{2 \cdot 1}$

Step 5.3.5.2.6.2

Multiply $2$ by $1$ .

$u = \frac{- 4 \pm 4 i \sqrt{2}}{2}$

Step 5.3.5.2.6.3

Simplify $\frac{- 4 \pm 4 i \sqrt{2}}{2}$ .

$u = - 2 \pm 2 i \sqrt{2}$

Step 5.3.5.2.6.4

Change the $\pm$ to $-$ .

$u = - 2 - 2 i \sqrt{2}$

Step 5.3.5.2.7

The final answer is the combination of both solutions.

$u = - 2 + 2 i \sqrt{2}, - 2 - 2 i \sqrt{2}$

Step 5.3.5.2.8

Substitute the real value of $u = {y'}^{2}$ back into the solved equation.

${y'}^{2} = - 2 + 2 i \sqrt{2}$

${({y'}^{2})}^{1} = - 2 - 2 i \sqrt{2}$

Step 5.3.5.2.9

Solve the first equation for $y'$ .

${y'}^{2} = - 2 + 2 i \sqrt{2}$

Step 5.3.5.2.10

Solve the equation for $y'$ .

Tap for more steps...

Step 5.3.5.2.10.1

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

$y' = \pm \sqrt{- 2 + 2 i \sqrt{2}}$

Step 5.3.5.2.10.2

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

Tap for more steps...

Step 5.3.5.2.10.2.1

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

$y' = \sqrt{- 2 + 2 i \sqrt{2}}$

Step 5.3.5.2.10.2.2

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

$y' = - \sqrt{- 2 + 2 i \sqrt{2}}$

Step 5.3.5.2.10.2.3

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

$y' = \sqrt{- 2 + 2 i \sqrt{2}}, - \sqrt{- 2 + 2 i \sqrt{2}}$

Step 5.3.5.2.11

Solve the second equation for $y'$ .

${({y'}^{2})}^{1} = - 2 - 2 i \sqrt{2}$

Step 5.3.5.2.12

Solve the equation for $y'$ .

Tap for more steps...

Step 5.3.5.2.12.1

Remove parentheses.

${y'}^{2} = - 2 - 2 i \sqrt{2}$

Step 5.3.5.2.12.2

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

$y' = \pm \sqrt{- 2 - 2 i \sqrt{2}}$

Step 5.3.5.2.12.3

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

Tap for more steps...

Step 5.3.5.2.12.3.1

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

$y' = \sqrt{- 2 - 2 i \sqrt{2}}$

Step 5.3.5.2.12.3.2

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

$y' = - \sqrt{- 2 - 2 i \sqrt{2}}$

Step 5.3.5.2.12.3.3

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

$y' = \sqrt{- 2 - 2 i \sqrt{2}}, - \sqrt{- 2 - 2 i \sqrt{2}}$

Step 5.3.5.2.13

The solution to ${y'}^{4} + 4 {y'}^{2} + 12 = 0$ is $y' = \sqrt{- 2 + 2 i \sqrt{2}}, - \sqrt{- 2 + 2 i \sqrt{2}}, \sqrt{- 2 - 2 i \sqrt{2}}, - \sqrt{- 2 - 2 i \sqrt{2}}$ .

$y' = \sqrt{- 2 + 2 i \sqrt{2}}, - \sqrt{- 2 + 2 i \sqrt{2}}, \sqrt{- 2 - 2 i \sqrt{2}}, - \sqrt{- 2 - 2 i \sqrt{2}}$

Step 5.3.6

The final solution is all the values that make $y' ({y'}^{4} + 4 {y'}^{2} + 12) = 0$ true.

$y' = 0, \sqrt{- 2 + 2 i \sqrt{2}}, - \sqrt{- 2 + 2 i \sqrt{2}}, \sqrt{- 2 - 2 i \sqrt{2}}, - \sqrt{- 2 - 2 i \sqrt{2}}$

Step 6

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

$\frac{d y}{d v} = 0, \sqrt{- 2 + 2 i \sqrt{2}}, - \sqrt{- 2 + 2 i \sqrt{2}}, \sqrt{- 2 - 2 i \sqrt{2}}, - \sqrt{- 2 - 2 i \sqrt{2}}$