Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x + 3$ and $g (x) = y$ .

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

Step 2.2

Differentiate.

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Step 2.2.1

By the Sum Rule, the derivative of $x + 3$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [3]$ .

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

Step 2.2.2

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

$\frac{y (1 + \frac{d}{d x} [3]) - (x + 3) \frac{d}{d x} [y]}{y^{2}}$

Step 2.2.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$\frac{y (1 + 0) - (x + 3) \frac{d}{d x} [y]}{y^{2}}$

Step 2.2.4

Simplify the expression.

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Step 2.2.4.1

Add $1$ and $0$ .

$\frac{y \cdot 1 - (x + 3) \frac{d}{d x} [y]}{y^{2}}$

Step 2.2.4.2

Multiply $y$ by $1$ .

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

Step 2.3

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

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Apply the distributive property.

$\frac{y - x y' - 1 \cdot 3 y'}{y^{2}}$

Step 2.4.3

Multiply $- 1$ by $3$ .

$\frac{y - x y' - 3 y'}{y^{2}}$

Step 2.4.4

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

$4 \cdot 1 + \frac{d}{d x} [y^{2}]$

Step 3.2.3

Multiply $4$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

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

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Step 3.3.1.1

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

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

Step 3.3.1.2

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

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

Step 3.3.1.3

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

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

Step 3.3.2

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

$4 + 2 y y'$

Step 3.4

Reorder terms.

$2 y y' + 4$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Multiply both sides by $y^{2}$ .

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

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 $\frac{- x y' + y - 3 y'}{y^{2}} y^{2}$ .

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Step 5.2.1.1.1

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

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Step 5.2.1.1.1.1

Cancel the common factor.

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

Step 5.2.1.1.1.2

Rewrite the expression.

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

Step 5.2.1.1.2

Reorder.

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Step 5.2.1.1.2.1

Move $x$ .

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

Step 5.2.1.1.2.2

Move $y$ .

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

Step 5.2.2

Simplify the right side.

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Step 5.2.2.1

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

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Step 5.2.2.1.1

Apply the distributive property.

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

Step 5.2.2.1.2

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

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Step 5.2.2.1.2.1

Move $y^{2}$ .

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

Step 5.2.2.1.2.2

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

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Step 5.2.2.1.2.2.1

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

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

Step 5.2.2.1.2.2.2

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

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

Step 5.2.2.1.2.3

Add $2$ and $1$ .

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

Step 5.2.2.1.3

Move $y^{3}$ .

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

Step 5.3

Solve for $y'$ .

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Step 5.3.1

Rewrite $- 1 y'$ as $- y'$ .

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

Step 5.3.2

Subtract $2 y' y^{3}$ from both sides of the equation.

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

Step 5.3.3

Subtract $y$ from both sides of the equation.

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

Step 5.3.4

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

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Step 5.3.4.1

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

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

Step 5.3.4.2

Factor $y'$ out of $- 3 y'$ .

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

Step 5.3.4.3

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

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

Step 5.3.4.4

Factor $y'$ out of $y' (- 1 x) + y' \cdot - 3$ .

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

Step 5.3.4.5

Factor $y'$ out of $y' (- 1 x - 3) + y' (- 2 y^{3})$ .

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

Step 5.3.5

Rewrite $- 1 x$ as $- x$ .

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

Step 5.3.6

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

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Step 5.3.6.1

Divide each term in $y' (- x - 3 - 2 y^{3}) = 4 y^{2} - y$ by $- x - 3 - 2 y^{3}$ .

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

Step 5.3.6.2

Simplify the left side.

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Step 5.3.6.2.1

Cancel the common factor of $- x - 3 - 2 y^{3}$ .

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Step 5.3.6.2.1.1

Cancel the common factor.

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

Step 5.3.6.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.6.3

Simplify the right side.

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Step 5.3.6.3.1

Combine the numerators over the common denominator.

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

Step 5.3.6.3.2

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

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Step 5.3.6.3.2.1

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

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

Step 5.3.6.3.2.2

Factor $y$ out of $- y$ .

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

Step 5.3.6.3.2.3

Factor $y$ out of $y (4 y) + y \cdot - 1$ .

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

Step 5.3.6.3.3

Factor $- 1$ out of $- x$ .

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

Step 5.3.6.3.4

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 5.3.6.3.5

Factor $- 1$ out of $- (x) - 1 (3)$ .

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

Step 5.3.6.3.6

Factor $- 1$ out of $- 2 y^{3}$ .

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

Step 5.3.6.3.7

Factor $- 1$ out of $- (x + 3) - (2 y^{3})$ .

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

Step 5.3.6.3.8

Rewrite negatives.

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Step 5.3.6.3.8.1

Rewrite $- (x + 3 + 2 y^{3})$ as $- 1 (x + 3 + 2 y^{3})$ .

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

Step 5.3.6.3.8.2

Move the negative in front of the fraction.

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

Step 6

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

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