Calculus Examples

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

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{5 x + y}$ as ${(5 x + y)}^{\frac{1}{2}}$ .

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

Step 2

Differentiate both sides of the equation.

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

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 $g (x) = 5 x + y$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $5 x + y$ .

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

Step 3.1.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [5 x + y]$

Step 3.1.3

Replace all occurrences of $u$ with $5 x + y$ .

$\frac{1}{2} {(5 x + y)}^{\frac{1}{2} - 1} \frac{d}{d x} [5 x + y]$

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Combine the numerators over the common denominator.

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

Step 3.5

Simplify the numerator.

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Step 3.5.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(5 x + y)}^{\frac{1 - 2}{2}} \frac{d}{d x} [5 x + y]$

Step 3.5.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(5 x + y)}^{\frac{- 1}{2}} \frac{d}{d x} [5 x + y]$

Step 3.6

Combine fractions.

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Step 3.6.1

Move the negative in front of the fraction.

$\frac{1}{2} {(5 x + y)}^{- \frac{1}{2}} \frac{d}{d x} [5 x + y]$

Step 3.6.2

Combine $\frac{1}{2}$ and ${(5 x + y)}^{- \frac{1}{2}}$ .

$\frac{{(5 x + y)}^{- \frac{1}{2}}}{2} \frac{d}{d x} [5 x + y]$

Step 3.6.3

Move ${(5 x + y)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}} \frac{d}{d x} [5 x + y]$

Step 3.7

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

$\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}} (\frac{d}{d x} [5 x] + \frac{d}{d x} [y])$

Step 3.8

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

$\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}} (5 \frac{d}{d x} [x] + \frac{d}{d x} [y])$

Step 3.9

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

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

Step 3.10

Multiply $5$ by $1$ .

$\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}} (5 + \frac{d}{d x} [y])$

Step 3.11

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

$\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}} (5 + y')$

Step 3.12

Simplify.

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Step 3.12.1

Reorder the factors of $\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}} (5 + y')$ .

$(5 + y') \frac{1}{2 {(5 x + y)}^{\frac{1}{2}}}$

Step 3.12.2

Multiply $5 + y'$ by $\frac{1}{2 {(5 x + y)}^{\frac{1}{2}}}$ .

$\frac{5 + y'}{2 {(5 x + y)}^{\frac{1}{2}}}$

Step 4

Differentiate the right side of the equation.

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Step 4.1

Differentiate.

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Step 4.1.1

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

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

Step 4.1.2

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$0 + \frac{d}{d x} [x^{2} y^{2}]$

Step 4.2

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

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Step 4.2.1

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

$0 + x^{2} \frac{d}{d x} [y^{2}] + y^{2} \frac{d}{d x} [x^{2}]$

Step 4.2.2

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

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Step 4.2.2.1

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

$0 + x^{2} (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{2}]$

Step 4.2.2.2

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

$0 + x^{2} (2 u \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{2}]$

Step 4.2.2.3

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

$0 + x^{2} (2 y \frac{d}{d x} [y]) + y^{2} \frac{d}{d x} [x^{2}]$

Step 4.2.3

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

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

Step 4.2.4

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

$0 + x^{2} (2 y y') + y^{2} (2 x)$

Step 4.2.5

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

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

Step 4.2.6

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

$0 + 2 x^{2} y y' + 2 y^{2} x$

Step 4.3

Add $0$ and $2 x^{2} y y' + 2 y^{2} x$ .

$2 x^{2} y y' + 2 y^{2} x$

Step 5

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

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

Step 6

Solve for $y'$ .

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Step 6.1

Multiply both sides by $2 {(5 x + y)}^{\frac{1}{2}}$ .

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

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

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Step 6.2.1.1.1

Rewrite using the commutative property of multiplication.

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

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.

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

Step 6.2.1.1.2.2

Rewrite the expression.

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

Step 6.2.1.1.3

Cancel the common factor of ${(5 x + y)}^{\frac{1}{2}}$ .

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Step 6.2.1.1.3.1

Cancel the common factor.

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

Step 6.2.1.1.3.2

Rewrite the expression.

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

Step 6.2.1.1.4

Reorder $5$ and $y'$ .

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

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

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

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Step 6.2.2.1.1

Apply the distributive property.

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

Step 6.2.2.1.2

Simplify the expression.

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Step 6.2.2.1.2.1

Multiply $2$ by $2$ .

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

Step 6.2.2.1.2.2

Multiply $2$ by $2$ .

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

Step 6.2.2.1.2.3

Move $y$ .

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

Step 6.2.2.1.2.4

Move $x^{2}$ .

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

Step 6.2.2.1.2.5

Move $y^{2}$ .

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

Step 6.3

Solve for $y'$ .

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Step 6.3.1

Subtract $4 y' x^{2} y {(5 x + y)}^{\frac{1}{2}}$ from both sides of the equation.

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

Step 6.3.2

Subtract $5$ from both sides of the equation.

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

Step 6.3.3

Factor $y'$ out of $y' - 4 y' x^{2} y {(5 x + y)}^{\frac{1}{2}}$ .

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Step 6.3.3.1

Factor $y'$ out of ${y'}^{1}$ .

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

Step 6.3.3.2

Factor $y'$ out of $- 4 y' x^{2} y {(5 x + y)}^{\frac{1}{2}}$ .

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

Step 6.3.3.3

Factor $y'$ out of $y' \cdot 1 + y' (- 4 x^{2} y {(5 x + y)}^{\frac{1}{2}})$ .

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

Step 6.3.4

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

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Step 6.3.4.1

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

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

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

Divide $y'$ by $1$ .

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

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.

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

Step 7

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

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