Calculus Examples

$4 x^{3} + \ln (y^{2}) + 2 y = 2 x$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $4 x^{3} + \ln (y^{2}) + 2 y$ with respect to $x$ is $\frac{d}{d x} [4 x^{3}] + \frac{d}{d x} [\ln (y^{2})] + \frac{d}{d x} [2 y]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [4 x^{3}]$ .

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Step 2.2.1

Since $4$ is constant with respect to $x$ , the derivative of $4 x^{3}$ with respect to $x$ is $4 \frac{d}{d x} [x^{3}]$ .

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $3$ by $4$ .

$12 x^{2} + \frac{d}{d x} [\ln (y^{2})] + \frac{d}{d x} [2 y]$

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

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

$12 x^{2} + \frac{d}{d u_{1}} [\ln (u_{1})] \frac{d}{d x} [y^{2}] + \frac{d}{d x} [2 y]$

Step 2.3.1.2

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

$12 x^{2} + \frac{1}{u_{1}} \frac{d}{d x} [y^{2}] + \frac{d}{d x} [2 y]$

Step 2.3.1.3

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

$12 x^{2} + \frac{1}{y^{2}} \frac{d}{d x} [y^{2}] + \frac{d}{d x} [2 y]$

Step 2.3.2

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

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Step 2.3.2.1

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

$12 x^{2} + \frac{1}{y^{2}} (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [2 y]$

Step 2.3.2.2

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

$12 x^{2} + \frac{1}{y^{2}} (2 u_{2} \frac{d}{d x} [y]) + \frac{d}{d x} [2 y]$

Step 2.3.2.3

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

$12 x^{2} + \frac{1}{y^{2}} (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [2 y]$

Step 2.3.3

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

$12 x^{2} + \frac{1}{y^{2}} (2 y y') + \frac{d}{d x} [2 y]$

Step 2.3.4

Combine $2$ and $\frac{1}{y^{2}}$ .

$12 x^{2} + \frac{2}{y^{2}} (y y') + \frac{d}{d x} [2 y]$

Step 2.3.5

Combine $y$ and $\frac{2}{y^{2}}$ .

$12 x^{2} + \frac{y \cdot 2}{y^{2}} y' + \frac{d}{d x} [2 y]$

Step 2.3.6

Combine $\frac{y \cdot 2}{y^{2}}$ and $y'$ .

$12 x^{2} + \frac{y \cdot 2 y'}{y^{2}} + \frac{d}{d x} [2 y]$

Step 2.3.7

Move $2$ to the left of $y$ .

$12 x^{2} + \frac{2 \cdot y y'}{y^{2}} + \frac{d}{d x} [2 y]$

Step 2.3.8

Cancel the common factor of $y$ and $y^{2}$ .

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Step 2.3.8.1

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

$12 x^{2} + \frac{y (2 y')}{y^{2}} + \frac{d}{d x} [2 y]$

Step 2.3.8.2

Cancel the common factors.

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Step 2.3.8.2.1

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

$12 x^{2} + \frac{y (2 y')}{y \cdot y} + \frac{d}{d x} [2 y]$

Step 2.3.8.2.2

Cancel the common factor.

$12 x^{2} + \frac{y (2 y')}{y \cdot y} + \frac{d}{d x} [2 y]$

Step 2.3.8.2.3

Rewrite the expression.

$12 x^{2} + \frac{2 y'}{y} + \frac{d}{d x} [2 y]$

Step 2.4

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

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Step 2.4.1

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

$12 x^{2} + \frac{2 y'}{y} + 2 \frac{d}{d x} [y]$

Step 2.4.2

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

$12 x^{2} + \frac{2 y'}{y} + 2 y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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 3.2

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 3.3

Multiply $2$ by $1$ .

$2$

Step 4

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

$12 x^{2} + \frac{2 y'}{y} + 2 y' = 2$

Step 5

Solve for $y'$ .

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Step 5.1

Find the LCD of the terms in the equation.

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Step 5.1.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$1, y, 1, 1$

Step 5.1.2

The LCM of one and any expression is the expression.

$y$

Step 5.2

Multiply each term in $12 x^{2} + \frac{2 y'}{y} + 2 y' = 2$ by $y$ to eliminate the fractions.

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Step 5.2.1

Multiply each term in $12 x^{2} + \frac{2 y'}{y} + 2 y' = 2$ by $y$ .

$12 x^{2} y + \frac{2 y'}{y} y + 2 y' y = 2 y$

Step 5.2.2

Simplify the left side.

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Step 5.2.2.1

Cancel the common factor of $y$ .

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Step 5.2.2.1.1

Cancel the common factor.

$12 x^{2} y + \frac{2 y'}{y} y + 2 y' y = 2 y$

Step 5.2.2.1.2

Rewrite the expression.

$12 x^{2} y + 2 y' + 2 y' y = 2 y$

Step 5.3

Solve the equation.

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Step 5.3.1

Subtract $12 x^{2} y$ from both sides of the equation.

$2 y' + 2 y' y = 2 y - 12 x^{2} y$

Step 5.3.2

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

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Step 5.3.2.1

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

$2 y' \cdot 1 + 2 y' y = 2 y - 12 x^{2} y$

Step 5.3.2.2

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

$2 y' \cdot 1 + 2 y' y = 2 y - 12 x^{2} y$

Step 5.3.2.3

Factor $2 y'$ out of $2 y' \cdot 1 + 2 y' y$ .

$2 y' (1 + y) = 2 y - 12 x^{2} y$

Step 5.3.3

Divide each term in $2 y' (1 + y) = 2 y - 12 x^{2} y$ by $2 (1 + y)$ and simplify.

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Step 5.3.3.1

Divide each term in $2 y' (1 + y) = 2 y - 12 x^{2} y$ by $2 (1 + y)$ .

$\frac{2 y' (1 + y)}{2 (1 + y)} = \frac{2 y}{2 (1 + y)} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.2

Simplify the left side.

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Step 5.3.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.3.2.1.1

Cancel the common factor.

$\frac{2 y' (1 + y)}{2 (1 + y)} = \frac{2 y}{2 (1 + y)} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.2.1.2

Rewrite the expression.

$\frac{y' (1 + y)}{1 + y} = \frac{2 y}{2 (1 + y)} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.2.2

Cancel the common factor of $1 + y$ .

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Step 5.3.3.2.2.1

Cancel the common factor.

$\frac{y' (1 + y)}{1 + y} = \frac{2 y}{2 (1 + y)} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{2 y}{2 (1 + y)} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.3

Simplify the right side.

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Step 5.3.3.3.1

Simplify each term.

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Step 5.3.3.3.1.1

Cancel the common factor of $2$ .

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Step 5.3.3.3.1.1.1

Cancel the common factor.

$y' = \frac{2 y}{2 (1 + y)} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.3.1.1.2

Rewrite the expression.

$y' = \frac{y}{1 + y} + \frac{- 12 x^{2} y}{2 (1 + y)}$

Step 5.3.3.3.1.2

Cancel the common factor of $- 12$ and $2$ .

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Step 5.3.3.3.1.2.1

Factor $2$ out of $- 12 x^{2} y$ .

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

Step 5.3.3.3.1.2.2

Cancel the common factors.

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Step 5.3.3.3.1.2.2.1

Cancel the common factor.

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

Step 5.3.3.3.1.2.2.2

Rewrite the expression.

$y' = \frac{y}{1 + y} + \frac{- 6 x^{2} y}{1 + y}$

Step 5.3.3.3.1.3

Move the negative in front of the fraction.

$y' = \frac{y}{1 + y} - \frac{6 x^{2} y}{1 + y}$

Step 5.3.3.3.2

Simplify terms.

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Step 5.3.3.3.2.1

Combine the numerators over the common denominator.

$y' = \frac{y - 6 x^{2} y}{1 + y}$

Step 5.3.3.3.2.2

Factor $y$ out of $y - 6 x^{2} y$ .

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Step 5.3.3.3.2.2.1

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

$y' = \frac{y^{1} - 6 x^{2} y}{1 + y}$

Step 5.3.3.3.2.2.2

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

$y' = \frac{y \cdot 1 - 6 x^{2} y}{1 + y}$

Step 5.3.3.3.2.2.3

Factor $y$ out of $- 6 x^{2} y$ .

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

Step 5.3.3.3.2.2.4

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

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

Step 6

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

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