Calculus Examples

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

Step 1

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

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

Step 2

Differentiate both sides of the equation.

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

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 $g (x) = 2 x + y$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $2 x + y$ .

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

Step 3.1.2

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

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

Step 3.1.3

Replace all occurrences of $u$ with $2 x + y$ .

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

Combine the numerators over the common denominator.

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

Step 3.5

Simplify the numerator.

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Step 3.5.1

Multiply $- 1$ by $2$ .

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

Step 3.5.2

Subtract $2$ from $1$ .

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

Step 3.6

Combine fractions.

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Step 3.6.1

Move the negative in front of the fraction.

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

Step 3.6.2

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

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

Step 3.6.3

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

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

Step 3.7

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

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

Step 3.8

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

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

Step 3.9

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

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

Step 3.10

Multiply $2$ by $1$ .

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

Step 3.11

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

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

Step 3.12

Simplify.

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Step 3.12.1

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

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

Step 3.12.2

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

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

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 $2 x$ with respect to $x$ is $2 \frac{d}{d x} [x]$ .

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

Step 4.2

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

$2 \cdot 1$

Step 4.3

Multiply $2$ by $1$ .

$2$

Step 5

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

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

Step 6

Solve for $y'$ .

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Step 6.1

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

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

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

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Step 6.2.1.1.1

Rewrite using the commutative property of multiplication.

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

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.

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

Step 6.2.1.1.2.2

Rewrite the expression.

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

Step 6.2.1.1.3

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

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Step 6.2.1.1.3.1

Cancel the common factor.

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

Step 6.2.1.1.3.2

Rewrite the expression.

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

Step 6.2.1.1.4

Reorder $2$ and $y'$ .

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

Step 6.2.2

Simplify the right side.

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Step 6.2.2.1

Multiply $2$ by $2$ .

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

Step 6.3

Subtract $2$ from both sides of the equation.

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

Step 7

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

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