Calculus Examples

$\sqrt{x^{3} + y^{3}} = 2 y$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{3} + y^{3}}$ as ${(x^{3} + y^{3})}^{\frac{1}{2}}$ .

${(x^{3} + y^{3})}^{\frac{1}{2}} = 2 y$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} ({(x^{3} + y^{3})}^{\frac{1}{2}}) = \frac{d}{d x} (2 y)$

Step 3

Differentiate the left side of the equation.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x^{3} + y^{3}$ .

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Step 3.1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{\frac{1}{2}}] \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.1.2

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

$\frac{1}{2} {u_{1}}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.1.3

Replace all occurrences of $u_{1}$ with $x^{3} + y^{3}$ .

$\frac{1}{2} {(x^{3} + y^{3})}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.2

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} {(x^{3} + y^{3})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.3

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} {(x^{3} + y^{3})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.4

Combine the numerators over the common denominator.

$\frac{1}{2} {(x^{3} + y^{3})}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.5

Simplify the numerator.

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Step 3.5.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(x^{3} + y^{3})}^{\frac{1 - 2}{2}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.5.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(x^{3} + y^{3})}^{\frac{- 1}{2}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.6

Differentiate.

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Step 3.6.1

Move the negative in front of the fraction.

$\frac{1}{2} {(x^{3} + y^{3})}^{- \frac{1}{2}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.6.2

Combine fractions.

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Step 3.6.2.1

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

$\frac{{(x^{3} + y^{3})}^{- \frac{1}{2}}}{2} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.6.2.2

Move ${(x^{3} + y^{3})}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} \frac{d}{d x} [x^{3} + y^{3}]$

Step 3.6.3

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

$\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (\frac{d}{d x} [x^{3}] + \frac{d}{d x} [y^{3}])$

Step 3.6.4

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

$\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (3 x^{2} + \frac{d}{d x} [y^{3}])$

Step 3.7

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.7.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

$\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (3 x^{2} + \frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [y])$

Step 3.7.2

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

$\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (3 x^{2} + 3 {u_{2}}^{2} \frac{d}{d x} [y])$

Step 3.7.3

Replace all occurrences of $u_{2}$ with $y$ .

$\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (3 x^{2} + 3 y^{2} \frac{d}{d x} [y])$

Step 3.8

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

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

Step 3.9

Simplify.

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Step 3.9.1

Reorder the factors of $\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (3 x^{2} + 3 y^{2} y')$ .

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

Step 3.9.2

Multiply $3 x^{2} + 3 y^{2} y'$ by $\frac{1}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}}$ .

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

Step 3.9.3

Factor $3$ out of $3 x^{2} + 3 y^{2} y'$ .

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Step 3.9.3.1

Factor $3$ out of $3 x^{2}$ .

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

Step 3.9.3.2

Factor $3$ out of $3 y^{2} y'$ .

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

Step 3.9.3.3

Factor $3$ out of $3 (x^{2}) + 3 (y^{2} y')$ .

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

Step 4

Differentiate the right side of the equation.

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Step 4.1

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

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

Step 4.2

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

$2 y'$

Step 5

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

$\frac{3 (x^{2} + y^{2} y')}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} = 2 y'$

Step 6

Solve for $y'$ .

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Step 6.1

Multiply both sides by $2 {(x^{3} + y^{3})}^{\frac{1}{2}}$ .

$\frac{3 (x^{2} + y^{2} y')}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (2 {(x^{3} + y^{3})}^{\frac{1}{2}}) = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2

Simplify.

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Step 6.2.1

Simplify the left side.

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Step 6.2.1.1

Simplify $\frac{3 (x^{2} + y^{2} y')}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$ .

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Step 6.2.1.1.1

Rewrite using the commutative property of multiplication.

$2 \frac{3 (x^{2} + y^{2} y')}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} {(x^{3} + y^{3})}^{\frac{1}{2}} = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.2

Cancel the common factor of $2$ .

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Step 6.2.1.1.2.1

Cancel the common factor.

$2 \frac{3 (x^{2} + y^{2} y')}{2 {(x^{3} + y^{3})}^{\frac{1}{2}}} {(x^{3} + y^{3})}^{\frac{1}{2}} = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.2.2

Rewrite the expression.

$\frac{3 (x^{2} + y^{2} y')}{{(x^{3} + y^{3})}^{\frac{1}{2}}} {(x^{3} + y^{3})}^{\frac{1}{2}} = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.3

Cancel the common factor of ${(x^{3} + y^{3})}^{\frac{1}{2}}$ .

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Step 6.2.1.1.3.1

Cancel the common factor.

$\frac{3 (x^{2} + y^{2} y')}{{(x^{3} + y^{3})}^{\frac{1}{2}}} {(x^{3} + y^{3})}^{\frac{1}{2}} = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.3.2

Rewrite the expression.

$3 (x^{2} + y^{2} y') = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.4

Apply the distributive property.

$3 x^{2} + 3 (y^{2} y') = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.5

Simplify the expression.

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Step 6.2.1.1.5.1

Move $y^{2}$ .

$3 x^{2} + 3 y' y^{2} = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.1.1.5.2

Reorder $3 x^{2}$ and $3 y' y^{2}$ .

$3 y' y^{2} + 3 x^{2} = 2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Simplify $2 y' (2 {(x^{3} + y^{3})}^{\frac{1}{2}})$ .

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Step 6.2.2.1.1

Rewrite using the commutative property of multiplication.

$3 y' y^{2} + 3 x^{2} = 2 \cdot 2 y' {(x^{3} + y^{3})}^{\frac{1}{2}}$

Step 6.2.2.1.2

Multiply $2$ by $2$ .

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

Step 6.3

Solve for $y'$ .

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Step 6.3.1

Subtract $4 y' {(x^{3} + y^{3})}^{\frac{1}{2}}$ from both sides of the equation.

$3 y' y^{2} + 3 x^{2} - 4 y' {(x^{3} + y^{3})}^{\frac{1}{2}} = 0$

Step 6.3.2

Subtract $3 x^{2}$ from both sides of the equation.

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

Step 6.3.3

Factor $y'$ out of $3 y' y^{2} - 4 y' {(x^{3} + y^{3})}^{\frac{1}{2}}$ .

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Step 6.3.3.1

Factor $y'$ out of $3 y' y^{2}$ .

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

Step 6.3.3.2

Factor $y'$ out of $- 4 y' {(x^{3} + y^{3})}^{\frac{1}{2}}$ .

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

Step 6.3.3.3

Factor $y'$ out of $y' (3 y^{2}) + y' (- 4 {(x^{3} + y^{3})}^{\frac{1}{2}})$ .

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

Step 6.3.4

Divide each term in $y' (3 y^{2} - 4 {(x^{3} + y^{3})}^{\frac{1}{2}}) = - 3 x^{2}$ by $3 y^{2} - 4 {(x^{3} + y^{3})}^{\frac{1}{2}}$ and simplify.

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Step 6.3.4.1

Divide each term in $y' (3 y^{2} - 4 {(x^{3} + y^{3})}^{\frac{1}{2}}) = - 3 x^{2}$ by $3 y^{2} - 4 {(x^{3} + y^{3})}^{\frac{1}{2}}$ .

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

Step 6.3.4.2

Simplify the left side.

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Step 6.3.4.2.1

Cancel the common factor.

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

Step 6.3.4.2.2

Divide $y'$ by $1$ .

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

Step 6.3.4.3

Simplify the right side.

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Step 6.3.4.3.1

Move the negative in front of the fraction.

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

Step 7

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

$\frac{d y}{d x} = - \frac{3 x^{2}}{3 y^{2} - 4 {(x^{3} + y^{3})}^{\frac{1}{2}}}$