Algebra Examples

$\ln (x + 2 y) = x^{4}$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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Step 2.1.1

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

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

Step 2.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

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

Step 2.1.3

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

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

Step 2.2

Differentiate.

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

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

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

Step 2.3

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

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

Step 2.4

Simplify.

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Step 2.4.1

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

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

Step 2.4.2

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

$\frac{1 + 2 y'}{x + 2 y}$

Step 3

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

$4 x^{3}$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Multiply both sides by $x + 2 y$ .

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

Step 5.2

Simplify.

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Step 5.2.1

Simplify the left side.

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Step 5.2.1.1

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

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Step 5.2.1.1.1

Cancel the common factor of $x + 2 y$ .

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Step 5.2.1.1.1.1

Cancel the common factor.

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

Step 5.2.1.1.1.2

Rewrite the expression.

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

Step 5.2.1.1.2

Reorder $1$ and $2 y'$ .

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

Simplify $4 x^{3} (x + 2 y)$ .

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Step 5.2.2.1.1

Apply the distributive property.

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

Step 5.2.2.1.2

Multiply $x^{3}$ by $x$ by adding the exponents.

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Step 5.2.2.1.2.1

Move $x$ .

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

Step 5.2.2.1.2.2

Multiply $x$ by $x^{3}$ .

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Step 5.2.2.1.2.2.1

Raise $x$ to the power of $1$ .

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

Step 5.2.2.1.2.2.2

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

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

Step 5.2.2.1.2.3

Add $1$ and $3$ .

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

Step 5.2.2.1.3

Rewrite using the commutative property of multiplication.

$2 y' + 1 = 4 x^{4} + 4 \cdot 2 x^{3} y$

Step 5.2.2.1.4

Multiply $4$ by $2$ .

$2 y' + 1 = 4 x^{4} + 8 x^{3} y$

Step 5.2.2.1.5

Reorder $4 x^{4}$ and $8 x^{3} y$ .

$2 y' + 1 = 8 x^{3} y + 4 x^{4}$

Step 5.3

Solve for $y'$ .

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Step 5.3.1

Subtract $1$ from both sides of the equation.

$2 y' = 8 x^{3} y + 4 x^{4} - 1$

Step 5.3.2

Divide each term in $2 y' = 8 x^{3} y + 4 x^{4} - 1$ by $2$ and simplify.

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Step 5.3.2.1

Divide each term in $2 y' = 8 x^{3} y + 4 x^{4} - 1$ by $2$ .

$\frac{2 y'}{2} = \frac{8 x^{3} y}{2} + \frac{4 x^{4}}{2} + \frac{- 1}{2}$

Step 5.3.2.2

Simplify the left side.

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Step 5.3.2.2.1

Cancel the common factor of $2$ .

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Step 5.3.2.2.1.1

Cancel the common factor.

$\frac{2 y'}{2} = \frac{8 x^{3} y}{2} + \frac{4 x^{4}}{2} + \frac{- 1}{2}$

Step 5.3.2.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{8 x^{3} y}{2} + \frac{4 x^{4}}{2} + \frac{- 1}{2}$

Step 5.3.2.3

Simplify the right side.

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Step 5.3.2.3.1

Simplify each term.

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Step 5.3.2.3.1.1

Cancel the common factor of $8$ and $2$ .

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Step 5.3.2.3.1.1.1

Factor $2$ out of $8 x^{3} y$ .

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

Step 5.3.2.3.1.1.2

Cancel the common factors.

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Step 5.3.2.3.1.1.2.1

Factor $2$ out of $2$ .

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

Step 5.3.2.3.1.1.2.2

Cancel the common factor.

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

Step 5.3.2.3.1.1.2.3

Rewrite the expression.

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

Step 5.3.2.3.1.1.2.4

Divide $4 x^{3} y$ by $1$ .

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

Step 5.3.2.3.1.2

Cancel the common factor of $4$ and $2$ .

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Step 5.3.2.3.1.2.1

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

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

Step 5.3.2.3.1.2.2

Cancel the common factors.

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Step 5.3.2.3.1.2.2.1

Factor $2$ out of $2$ .

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

Step 5.3.2.3.1.2.2.2

Cancel the common factor.

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

Step 5.3.2.3.1.2.2.3

Rewrite the expression.

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

Step 5.3.2.3.1.2.2.4

Divide $2 x^{4}$ by $1$ .

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

Step 5.3.2.3.1.3

Move the negative in front of the fraction.

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

Step 6

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

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