Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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Step 2.2.1.1

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

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

Step 2.2.1.2

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

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

Step 2.2.1.3

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

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

Step 2.2.2

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

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

Step 2.3

Evaluate $\frac{d}{d x} [- 4 x y]$ .

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Step 2.3.1

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

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

Step 2.3.2

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 $f (x) = x$ and $g (x) = y$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $y$ by $1$ .

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

Step 2.4

The derivative of $\arctan (x)$ with respect to $x$ is $\frac{1}{1 + x^{2}}$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Combine terms.

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Step 2.5.2.1

To write $2 y y'$ as a fraction with a common denominator, multiply by $\frac{1 + x^{2}}{1 + x^{2}}$ .

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

Step 2.5.2.2

Combine the numerators over the common denominator.

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

Step 2.5.2.3

To write $- 4 x y'$ as a fraction with a common denominator, multiply by $\frac{1 + x^{2}}{1 + x^{2}}$ .

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

Step 2.5.2.4

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

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

Step 2.5.2.5

Combine the numerators over the common denominator.

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

Step 2.5.2.6

To write $- 4 y$ as a fraction with a common denominator, multiply by $\frac{1 + x^{2}}{1 + x^{2}}$ .

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

Step 2.5.2.7

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

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

Step 2.5.2.8

Combine the numerators over the common denominator.

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

Step 2.5.3

Reorder terms.

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

Step 3

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

$0$

Step 4

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

$\frac{- 4 y (1 + x^{2}) - 4 x (1 + x^{2}) y' + 2 y (1 + x^{2}) y' + 1}{x^{2} + 1} = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Set the numerator equal to zero.

$- 4 y (1 + x^{2}) - 4 x (1 + x^{2}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2

Solve the equation for $y'$ .

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Step 5.2.1

Simplify each term.

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Step 5.2.1.1

Apply the distributive property.

$- 4 y \cdot 1 - 4 y x^{2} - 4 x (1 + x^{2}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.2

Multiply $- 4$ by $1$ .

$- 4 y - 4 y x^{2} - 4 x (1 + x^{2}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.3

Apply the distributive property.

$- 4 y - 4 y x^{2} + (- 4 x \cdot 1 - 4 x \cdot x^{2}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.4

Multiply $- 4$ by $1$ .

$- 4 y - 4 y x^{2} + (- 4 x - 4 x \cdot x^{2}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.5

Multiply $x$ by $x^{2}$ by adding the exponents.

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Step 5.2.1.5.1

Move $x^{2}$ .

$- 4 y - 4 y x^{2} + (- 4 x - 4 (x^{2} x)) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.5.2

Multiply $x^{2}$ by $x$ .

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Step 5.2.1.5.2.1

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

$- 4 y - 4 y x^{2} + (- 4 x - 4 (x^{2} x^{1})) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.5.2.2

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

$- 4 y - 4 y x^{2} + (- 4 x - 4 x^{2 + 1}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.5.3

Add $2$ and $1$ .

$- 4 y - 4 y x^{2} + (- 4 x - 4 x^{3}) y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.6

Apply the distributive property.

$- 4 y - 4 y x^{2} - 4 x y' - 4 x^{3} y' + 2 y (1 + x^{2}) y' + 1 = 0$

Step 5.2.1.7

Apply the distributive property.

$- 4 y - 4 y x^{2} - 4 x y' - 4 x^{3} y' + (2 y \cdot 1 + 2 y x^{2}) y' + 1 = 0$

Step 5.2.1.8

Multiply $2$ by $1$ .

$- 4 y - 4 y x^{2} - 4 x y' - 4 x^{3} y' + (2 y + 2 y x^{2}) y' + 1 = 0$

Step 5.2.1.9

Apply the distributive property.

$- 4 y - 4 y x^{2} - 4 x y' - 4 x^{3} y' + 2 y y' + 2 y x^{2} y' + 1 = 0$

Step 5.2.2

Move all terms not containing $y'$ to the right side of the equation.

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Step 5.2.2.1

Add $4 y$ to both sides of the equation.

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

Step 5.2.2.2

Add $4 y x^{2}$ to both sides of the equation.

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

Step 5.2.2.3

Subtract $1$ from both sides of the equation.

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

Step 5.2.3

Factor $2 y'$ out of $- 4 x y' - 4 x^{3} y' + 2 y y' + 2 y x^{2} y'$ .

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Step 5.2.3.1

Factor $2 y'$ out of $- 4 x y'$ .

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

Step 5.2.3.2

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

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

Step 5.2.3.3

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

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

Step 5.2.3.4

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

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

Step 5.2.3.5

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

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

Step 5.2.3.6

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

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

Step 5.2.3.7

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

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

Step 5.2.4

Factor out the greatest common factor from each group.

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Step 5.2.4.1

Group the first two terms and the last two terms.

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

Step 5.2.4.2

Factor out the greatest common factor (GCF) from each group.

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

Step 5.2.5

Factor.

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Step 5.2.5.1

Factor the polynomial by factoring out the greatest common factor, $- 1 - x^{2}$ .

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

Step 5.2.5.2

Remove unnecessary parentheses.

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

Step 5.2.6

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

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Step 5.2.6.1

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

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

Step 5.2.6.2

Simplify the left side.

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Step 5.2.6.2.1

Cancel the common factor of $2$ .

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Step 5.2.6.2.1.1

Cancel the common factor.

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

Step 5.2.6.2.1.2

Rewrite the expression.

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

Step 5.2.6.2.2

Cancel the common factor of $- 1 - x^{2}$ .

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Step 5.2.6.2.2.1

Cancel the common factor.

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

Step 5.2.6.2.2.2

Rewrite the expression.

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

Step 5.2.6.2.3

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

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Step 5.2.6.2.3.1

Cancel the common factor.

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

Step 5.2.6.2.3.2

Divide $y'$ by $1$ .

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

Step 5.2.6.3

Simplify the right side.

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Step 5.2.6.3.1

Simplify each term.

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Step 5.2.6.3.1.1

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

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Step 5.2.6.3.1.1.1

Factor $2$ out of $4 y$ .

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

Step 5.2.6.3.1.1.2

Cancel the common factors.

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Step 5.2.6.3.1.1.2.1

Factor $2$ out of $2 (- 1 - x^{2}) (2 x - y)$ .

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

Step 5.2.6.3.1.1.2.2

Cancel the common factor.

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

Step 5.2.6.3.1.1.2.3

Rewrite the expression.

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

Step 5.2.6.3.1.2

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

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Step 5.2.6.3.1.2.1

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

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

Step 5.2.6.3.1.2.2

Cancel the common factors.

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Step 5.2.6.3.1.2.2.1

Factor $2$ out of $2 (- 1 - x^{2}) (2 x - y)$ .

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

Step 5.2.6.3.1.2.2.2

Cancel the common factor.

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

Step 5.2.6.3.1.2.2.3

Rewrite the expression.

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

Step 5.2.6.3.1.3

Move the negative in front of the fraction.

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

Step 6

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

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