Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

Differentiate.

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Step 2.1.1

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

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

Step 2.1.2

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

$0 + \frac{d}{d x} [x] + \frac{d}{d x} [- y^{2}]$

Step 2.1.3

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

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

Step 2.2

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

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Step 2.2.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{2}$ with respect to $x$ is $- \frac{d}{d x} [y^{2}]$ .

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

Step 2.2.2

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 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

$0 + 1 - (2 y y')$

Step 2.2.4

Multiply $2$ by $- 1$ .

$0 + 1 - 2 y y'$

Step 2.3

Simplify.

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Step 2.3.1

Add $0$ and $1$ .

$1 - 2 y y'$

Step 2.3.2

Reorder terms.

$- 2 y y' + 1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

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

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

Step 3.3

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

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

Step 3.4

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

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

Step 3.5

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

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

Step 3.6

Simplify.

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Step 3.6.1

Apply the distributive property.

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

Step 3.6.2

Multiply $3$ by $- 1$ .

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

Step 3.6.3

Reorder terms.

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

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

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

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

Step 5.2

Subtract $1$ from both sides of the equation.

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

Step 5.3

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

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

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

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

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

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

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

Step 5.4.3.2

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

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

Step 5.4.3.3

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

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

Step 5.4.3.4

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

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

Step 5.4.3.5

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

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

Step 5.4.3.6

Factor $- 1$ out of $- 2 y$ .

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

Step 5.4.3.7

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

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

Step 5.4.3.8

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

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

Step 5.4.3.9

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

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

Step 5.4.3.10

Cancel the common factor.

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

Step 5.4.3.11

Rewrite the expression.

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

Step 6

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

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