Calculus Examples

$x^{2} y^{3} - 2 x y = 6 x + y + 1$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{2} y^{3} - 2 x y) = \frac{d}{d x} (6 x + y + 1)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

Evaluate $\frac{d}{d x} [x^{2} y^{3}]$ .

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Step 2.2.1

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 $f (x) = x^{2}$ and $g (x) = y^{3}$ .

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

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

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Step 2.2.2.1

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

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

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 $n = 3$ .

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Move $3$ to the left of $x^{2}$ .

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

Step 2.2.6

Move $2$ to the left of $y^{3}$ .

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

Step 2.3

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

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Step 2.3.1

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

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

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 $g (x) = y$ .

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

Step 2.3.3

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

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

Step 2.3.4

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

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

Step 2.3.5

Multiply $y$ by $1$ .

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Remove unnecessary parentheses.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $6 x + y + 1$ with respect to $x$ is $\frac{d}{d x} [6 x] + \frac{d}{d x} [y] + \frac{d}{d x} [1]$ .

$\frac{d}{d x} [6 x] + \frac{d}{d x} [y] + \frac{d}{d x} [1]$

Step 3.2

Evaluate $\frac{d}{d x} [6 x]$ .

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Step 3.2.1

Since $6$ is constant with respect to $x$ , the derivative of $6 x$ with respect to $x$ is $6 \frac{d}{d x} [x]$ .

$6 \frac{d}{d x} [x] + \frac{d}{d x} [y] + \frac{d}{d x} [1]$

Step 3.2.2

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

$6 \cdot 1 + \frac{d}{d x} [y] + \frac{d}{d x} [1]$

Step 3.2.3

Multiply $6$ by $1$ .

$6 + \frac{d}{d x} [y] + \frac{d}{d x} [1]$

Step 3.3

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

$6 + y' + \frac{d}{d x} [1]$

Step 3.4

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$6 + y' + 0$

Step 3.5

Add $6 + y'$ and $0$ .

$6 + y'$

Step 4

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

$3 x^{2} y^{2} y' + 2 y^{3} x - 2 x y' - 2 y = 6 + y'$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $y'$ from both sides of the equation.

$3 x^{2} y^{2} y' + 2 y^{3} x - 2 x y' - 2 y - y' = 6$

Step 5.2

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

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Step 5.2.1

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

$3 x^{2} y^{2} y' - 2 x y' - 2 y - y' = 6 - 2 y^{3} x$

Step 5.2.2

Add $2 y$ to both sides of the equation.

$3 x^{2} y^{2} y' - 2 x y' - y' = 6 - 2 y^{3} x + 2 y$

Step 5.3

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

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Step 5.3.1

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

$y' (3 x^{2} y^{2}) - 2 x y' - y' = 6 - 2 y^{3} x + 2 y$

Step 5.3.2

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

$y' (3 x^{2} y^{2}) + y' (- 2 x) - y' = 6 - 2 y^{3} x + 2 y$

Step 5.3.3

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

$y' (3 x^{2} y^{2}) + y' (- 2 x) + y' \cdot - 1 = 6 - 2 y^{3} x + 2 y$

Step 5.3.4

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

$y' (3 x^{2} y^{2} - 2 x) + y' \cdot - 1 = 6 - 2 y^{3} x + 2 y$

Step 5.3.5

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

$y' (3 x^{2} y^{2} - 2 x - 1) = 6 - 2 y^{3} x + 2 y$

Step 5.4

Divide each term in $y' (3 x^{2} y^{2} - 2 x - 1) = 6 - 2 y^{3} x + 2 y$ by $3 x^{2} y^{2} - 2 x - 1$ and simplify.

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Step 5.4.1

Divide each term in $y' (3 x^{2} y^{2} - 2 x - 1) = 6 - 2 y^{3} x + 2 y$ by $3 x^{2} y^{2} - 2 x - 1$ .

$\frac{y' (3 x^{2} y^{2} - 2 x - 1)}{3 x^{2} y^{2} - 2 x - 1} = \frac{6}{3 x^{2} y^{2} - 2 x - 1} + \frac{- 2 y^{3} x}{3 x^{2} y^{2} - 2 x - 1} + \frac{2 y}{3 x^{2} y^{2} - 2 x - 1}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $3 x^{2} y^{2} - 2 x - 1$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (3 x^{2} y^{2} - 2 x - 1)}{3 x^{2} y^{2} - 2 x - 1} = \frac{6}{3 x^{2} y^{2} - 2 x - 1} + \frac{- 2 y^{3} x}{3 x^{2} y^{2} - 2 x - 1} + \frac{2 y}{3 x^{2} y^{2} - 2 x - 1}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{6}{3 x^{2} y^{2} - 2 x - 1} + \frac{- 2 y^{3} x}{3 x^{2} y^{2} - 2 x - 1} + \frac{2 y}{3 x^{2} y^{2} - 2 x - 1}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Move the negative in front of the fraction.

$y' = \frac{6}{3 x^{2} y^{2} - 2 x - 1} - \frac{2 y^{3} x}{3 x^{2} y^{2} - 2 x - 1} + \frac{2 y}{3 x^{2} y^{2} - 2 x - 1}$

Step 5.4.3.2

Combine the numerators over the common denominator.

$y' = \frac{6 - 2 y^{3} x}{3 x^{2} y^{2} - 2 x - 1} + \frac{2 y}{3 x^{2} y^{2} - 2 x - 1}$

Step 5.4.3.3

Combine the numerators over the common denominator.

$y' = \frac{6 - 2 y^{3} x + 2 y}{3 x^{2} y^{2} - 2 x - 1}$

Step 5.4.3.4

Factor $2$ out of $6 - 2 y^{3} x + 2 y$ .

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Step 5.4.3.4.1

Factor $2$ out of $6$ .

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

Step 5.4.3.4.2

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

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

Step 5.4.3.4.3

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

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

Step 5.4.3.4.4

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

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

Step 6

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

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