Calculus Examples

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

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (- x^{3} y - 2 y^{2} + 4 x) = \frac{d}{d x} (0)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

$\frac{d}{d x} [- x^{3} y] + \frac{d}{d x} [- 2 y^{2}] + \frac{d}{d x} [4 x]$

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3} y$ with respect to $x$ is $- \frac{d}{d x} [x^{3} y]$ .

$- \frac{d}{d x} [x^{3} y] + \frac{d}{d x} [- 2 y^{2}] + \frac{d}{d x} [4 x]$

Step 2.2.2

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 $f (x) = x^{3}$ and $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Move $3$ to the left of $y$ .

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

Step 2.3

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

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Step 2.3.1

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

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

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 $u$ as $y$ .

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

Step 2.3.2.2

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

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

Step 2.3.2.3

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

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

Step 2.3.3

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

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

Step 2.3.4

Multiply $2$ by $- 2$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $4$ by $1$ .

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

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

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

Step 2.5.2

Multiply $3$ by $- 1$ .

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

Step 2.5.3

Reorder terms.

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

Step 3

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

$0$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

Add $3 x^{2} y$ to both sides of the equation.

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

Step 5.1.2

Subtract $4$ from both sides of the equation.

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

Step 5.2

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

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Step 5.2.1

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

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

Step 5.2.2

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

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

Step 5.2.3

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

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $- x^{3} - 4 y$ .

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

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

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

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

Step 5.3.3.2

Factor $- 1$ out of $- x^{3}$ .

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

Step 5.3.3.3

Factor $- 1$ out of $- 4 y$ .

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

Step 5.3.3.4

Factor $- 1$ out of $- (x^{3}) - (4 y)$ .

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

Step 5.3.3.5

Rewrite negatives.

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Step 5.3.3.5.1

Rewrite $- (x^{3} + 4 y)$ as $- 1 (x^{3} + 4 y)$ .

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

Step 5.3.3.5.2

Move the negative in front of the fraction.

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

Step 6

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

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