Calculus Examples

$\sqrt{x^{4} + y^{2}} = 5 x + 2 y^{3}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{4} + y^{2}}$ as ${(x^{4} + y^{2})}^{\frac{1}{2}}$ .

${(x^{4} + y^{2})}^{\frac{1}{2}} = 5 x + 2 y^{3}$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} ({(x^{4} + y^{2})}^{\frac{1}{2}}) = \frac{d}{d x} (5 x + 2 y^{3})$

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 $g (x) = x^{4} + y^{2}$ .

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Step 3.1.1

To apply the Chain Rule, set $u_{1}$ as $x^{4} + y^{2}$ .

$\frac{d}{d u_{1}} [{u_{1}}^{\frac{1}{2}}] \frac{d}{d x} [x^{4} + y^{2}]$

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 $n = \frac{1}{2}$ .

$\frac{1}{2} {u_{1}}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{4} + y^{2}]$

Step 3.1.3

Replace all occurrences of $u_{1}$ with $x^{4} + y^{2}$ .

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

Step 3.2

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

$\frac{1}{2} {(x^{4} + y^{2})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d x} [x^{4} + y^{2}]$

Step 3.3

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

$\frac{1}{2} {(x^{4} + y^{2})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d x} [x^{4} + y^{2}]$

Step 3.4

Combine the numerators over the common denominator.

$\frac{1}{2} {(x^{4} + y^{2})}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d x} [x^{4} + y^{2}]$

Step 3.5

Simplify the numerator.

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Step 3.5.1

Multiply $- 1$ by $2$ .

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

Step 3.5.2

Subtract $2$ from $1$ .

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

Step 3.6

Differentiate.

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Step 3.6.1

Move the negative in front of the fraction.

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

Step 3.6.2

Combine fractions.

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Step 3.6.2.1

Combine $\frac{1}{2}$ and ${(x^{4} + y^{2})}^{- \frac{1}{2}}$ .

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

Step 3.6.2.2

Move ${(x^{4} + y^{2})}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(x^{4} + y^{2})}^{\frac{1}{2}}} \frac{d}{d x} [x^{4} + y^{2}]$

Step 3.6.3

By the Sum Rule, the derivative of $x^{4} + y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [y^{2}]$ .

$\frac{1}{2 {(x^{4} + y^{2})}^{\frac{1}{2}}} (\frac{d}{d x} [x^{4}] + \frac{d}{d x} [y^{2}])$

Step 3.6.4

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

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 $f (x) = x^{2}$ and $g (x) = y$ .

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Step 3.7.1

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

$\frac{1}{2 {(x^{4} + y^{2})}^{\frac{1}{2}}} (4 x^{3} + \frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y])$

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 $n = 2$ .

$\frac{1}{2 {(x^{4} + y^{2})}^{\frac{1}{2}}} (4 x^{3} + 2 u_{2} \frac{d}{d x} [y])$

Step 3.7.3

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

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

Step 3.8

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

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

Step 3.9

Simplify.

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Step 3.9.1

Reorder the factors of $\frac{1}{2 {(x^{4} + y^{2})}^{\frac{1}{2}}} (4 x^{3} + 2 y y')$ .

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

Step 3.9.2

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

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

Step 3.9.3

Factor $2$ out of $4 x^{3}$ .

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

Step 3.9.4

Factor $2$ out of $2 y y'$ .

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

Step 3.9.5

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

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

Step 3.9.6

Cancel the common factors.

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Step 3.9.6.1

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

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

Step 3.9.6.2

Cancel the common factor.

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

Step 3.9.6.3

Rewrite the expression.

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

Step 4

Differentiate the right side of the equation.

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Step 4.1

By the Sum Rule, the derivative of $5 x + 2 y^{3}$ with respect to $x$ is $\frac{d}{d x} [5 x] + \frac{d}{d x} [2 y^{3}]$ .

$\frac{d}{d x} [5 x] + \frac{d}{d x} [2 y^{3}]$

Step 4.2

Evaluate $\frac{d}{d x} [5 x]$ .

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Step 4.2.1

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

$5 \frac{d}{d x} [x] + \frac{d}{d x} [2 y^{3}]$

Step 4.2.2

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

$5 \cdot 1 + \frac{d}{d x} [2 y^{3}]$

Step 4.2.3

Multiply $5$ by $1$ .

$5 + \frac{d}{d x} [2 y^{3}]$

Step 4.3

Evaluate $\frac{d}{d x} [2 y^{3}]$ .

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Step 4.3.1

Since $2$ is constant with respect to $x$ , the derivative of $2 y^{3}$ with respect to $x$ is $2 \frac{d}{d x} [y^{3}]$ .

$5 + 2 \frac{d}{d x} [y^{3}]$

Step 4.3.2

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 4.3.2.1

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

$5 + 2 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

Step 4.3.2.2

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

$5 + 2 (3 u^{2} \frac{d}{d x} [y])$

Step 4.3.2.3

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

$5 + 2 (3 y^{2} \frac{d}{d x} [y])$

Step 4.3.3

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

$5 + 2 (3 y^{2} y')$

Step 4.3.4

Multiply $3$ by $2$ .

$5 + 6 y^{2} y'$

Step 4.4

Reorder terms.

$6 y^{2} y' + 5$

Step 5

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

$\frac{2 x^{3} + y y'}{{(x^{4} + y^{2})}^{\frac{1}{2}}} = 6 y^{2} y' + 5$

Step 6

Solve for $y'$ .

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Step 6.1

Multiply both sides by ${(x^{4} + y^{2})}^{\frac{1}{2}}$ .

$\frac{2 x^{3} + y y'}{{(x^{4} + y^{2})}^{\frac{1}{2}}} {(x^{4} + y^{2})}^{\frac{1}{2}} = (6 y^{2} y' + 5) {(x^{4} + y^{2})}^{\frac{1}{2}}$

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

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Step 6.2.1.1.1

Cancel the common factor of ${(x^{4} + y^{2})}^{\frac{1}{2}}$ .

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Step 6.2.1.1.1.1

Cancel the common factor.

$\frac{2 x^{3} + y y'}{{(x^{4} + y^{2})}^{\frac{1}{2}}} {(x^{4} + y^{2})}^{\frac{1}{2}} = (6 y^{2} y' + 5) {(x^{4} + y^{2})}^{\frac{1}{2}}$

Step 6.2.1.1.1.2

Rewrite the expression.

$2 x^{3} + y y' = (6 y^{2} y' + 5) {(x^{4} + y^{2})}^{\frac{1}{2}}$

Step 6.2.1.1.2

Reorder $y$ and $y'$ .

$2 x^{3} + y' y = (6 y^{2} y' + 5) {(x^{4} + y^{2})}^{\frac{1}{2}}$

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Simplify $(6 y^{2} y' + 5) {(x^{4} + y^{2})}^{\frac{1}{2}}$ .

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Step 6.2.2.1.1

Apply the distributive property.

$2 x^{3} + y' y = 6 y^{2} y' {(x^{4} + y^{2})}^{\frac{1}{2}} + 5 {(x^{4} + y^{2})}^{\frac{1}{2}}$

Step 6.2.2.1.2

Move $y^{2}$ .

$2 x^{3} + y' y = 6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}} + 5 {(x^{4} + y^{2})}^{\frac{1}{2}}$

Step 6.3

Solve for $y'$ .

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Step 6.3.1

Subtract $6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}}$ from both sides of the equation.

$2 x^{3} + y' y - 6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}} = 5 {(x^{4} + y^{2})}^{\frac{1}{2}}$

Step 6.3.2

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

$y' y - 6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}} = 5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}$

Step 6.3.3

Factor $y' y$ out of $y' y - 6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}}$ .

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Step 6.3.3.1

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

$y' y \cdot 1 - 6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}} = 5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}$

Step 6.3.3.2

Factor $y' y$ out of $- 6 y' y^{2} {(x^{4} + y^{2})}^{\frac{1}{2}}$ .

$y' y \cdot 1 + y' y (- 6 y {(x^{4} + y^{2})}^{\frac{1}{2}}) = 5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}$

Step 6.3.3.3

Factor $y' y$ out of $y' y \cdot 1 + y' y (- 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})$ .

$y' y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}}) = 5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}$

Step 6.3.4

Divide each term in $y' y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}}) = 5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}$ by $y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})$ and simplify.

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Step 6.3.4.1

Divide each term in $y' y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}}) = 5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}$ by $y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})$ .

$\frac{y' y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})} = \frac{5 {(x^{4} + y^{2})}^{\frac{1}{2}}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})} + \frac{- 2 x^{3}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}$

Step 6.3.4.2

Simplify the left side.

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Step 6.3.4.2.1

Cancel the common factor.

Step 6.3.4.2.2

Rewrite the expression.

$\frac{y' (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}{1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}}} = \frac{5 {(x^{4} + y^{2})}^{\frac{1}{2}}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})} + \frac{- 2 x^{3}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}$

Step 6.3.4.2.3

Cancel the common factor.

Step 6.3.4.2.4

Divide $y'$ by $1$ .

$y' = \frac{5 {(x^{4} + y^{2})}^{\frac{1}{2}}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})} + \frac{- 2 x^{3}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}$

Step 6.3.4.3

Simplify the right side.

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Step 6.3.4.3.1

Combine the numerators over the common denominator.

$y' = \frac{5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}$

Step 7

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

$\frac{d y}{d x} = \frac{5 {(x^{4} + y^{2})}^{\frac{1}{2}} - 2 x^{3}}{y (1 - 6 y {(x^{4} + y^{2})}^{\frac{1}{2}})}$