Calculus Examples

$3 x^{2} - 2 x y + 5 y^{2} = 1$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d y} (3 x^{2} - 2 x y + 5 y^{2}) = \frac{d}{d y} (1)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $3 x^{2} - 2 x y + 5 y^{2}$ with respect to $y$ is $\frac{d}{d y} [3 x^{2}] + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$ .

$\frac{d}{d y} [3 x^{2}] + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.2

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

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Step 2.2.1

Since $3$ is constant with respect to $y$ , the derivative of $3 x^{2}$ with respect to $y$ is $3 \frac{d}{d y} [x^{2}]$ .

$3 \frac{d}{d y} [x^{2}] + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.2.2

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

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Step 2.2.2.1

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

$3 (\frac{d}{d u} [u^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

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 $n = 2$ .

$3 (2 u \frac{d}{d y} [x]) + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.2.2.3

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

$3 (2 x \frac{d}{d y} [x]) + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.2.3

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

$3 (2 x x') + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.2.4

Multiply $2$ by $3$ .

$6 x x' + \frac{d}{d y} [- 2 x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.3

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

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Step 2.3.1

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

$6 x x' - 2 \frac{d}{d y} [x y] + \frac{d}{d y} [5 y^{2}]$

Step 2.3.2

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

$6 x x' - 2 (x \frac{d}{d y} [y] + y \frac{d}{d y} [x]) + \frac{d}{d y} [5 y^{2}]$

Step 2.3.3

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

$6 x x' - 2 (x \cdot 1 + y \frac{d}{d y} [x]) + \frac{d}{d y} [5 y^{2}]$

Step 2.3.4

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

$6 x x' - 2 (x \cdot 1 + y x') + \frac{d}{d y} [5 y^{2}]$

Step 2.3.5

Multiply $x$ by $1$ .

$6 x x' - 2 (x + y x') + \frac{d}{d y} [5 y^{2}]$

Step 2.4

Evaluate $\frac{d}{d y} [5 y^{2}]$ .

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Step 2.4.1

Since $5$ is constant with respect to $y$ , the derivative of $5 y^{2}$ with respect to $y$ is $5 \frac{d}{d y} [y^{2}]$ .

$6 x x' - 2 (x + y x') + 5 \frac{d}{d y} [y^{2}]$

Step 2.4.2

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

$6 x x' - 2 (x + y x') + 5 (2 y)$

Step 2.4.3

Multiply $2$ by $5$ .

$6 x x' - 2 (x + y x') + 10 y$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$6 x x' - 2 x - 2 (y x') + 10 y$

Step 2.5.2

Remove unnecessary parentheses.

$6 x x' - 2 x - 2 y x' + 10 y$

Step 2.5.3

Reorder terms.

$6 x x' - 2 y x' - 2 x + 10 y$

Step 3

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

$0$

Step 4

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

$6 x x' - 2 y x' - 2 x + 10 y = 0$

Step 5

Solve for $x'$ .

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Step 5.1

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

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Step 5.1.1

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

$6 x x' - 2 y x' + 10 y = 2 x$

Step 5.1.2

Subtract $10 y$ from both sides of the equation.

$6 x x' - 2 y x' = 2 x - 10 y$

Step 5.2

Factor $2 x'$ out of $6 x x' - 2 y x'$ .

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Step 5.2.1

Factor $2 x'$ out of $6 x x'$ .

$2 x' (3 x) - 2 y x' = 2 x - 10 y$

Step 5.2.2

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

$2 x' (3 x) + 2 x' (- y) = 2 x - 10 y$

Step 5.2.3

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

$2 x' (3 x - y) = 2 x - 10 y$

Step 5.3

Divide each term in $2 x' (3 x - y) = 2 x - 10 y$ by $2 (3 x - y)$ and simplify.

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Step 5.3.1

Divide each term in $2 x' (3 x - y) = 2 x - 10 y$ by $2 (3 x - y)$ .

$\frac{2 x' (3 x - y)}{2 (3 x - y)} = \frac{2 x}{2 (3 x - y)} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

Cancel the common factor of $2$ .

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{2 x' (3 x - y)}{2 (3 x - y)} = \frac{2 x}{2 (3 x - y)} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{x' (3 x - y)}{3 x - y} = \frac{2 x}{2 (3 x - y)} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.2.2

Cancel the common factor of $3 x - y$ .

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Step 5.3.2.2.1

Cancel the common factor.

$\frac{x' (3 x - y)}{3 x - y} = \frac{2 x}{2 (3 x - y)} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.2.2.2

Divide $x'$ by $1$ .

$x' = \frac{2 x}{2 (3 x - y)} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Simplify each term.

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Step 5.3.3.1.1

Cancel the common factor of $2$ .

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Step 5.3.3.1.1.1

Cancel the common factor.

$x' = \frac{2 x}{2 (3 x - y)} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.3.1.1.2

Rewrite the expression.

$x' = \frac{x}{3 x - y} + \frac{- 10 y}{2 (3 x - y)}$

Step 5.3.3.1.2

Cancel the common factor of $- 10$ and $2$ .

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Step 5.3.3.1.2.1

Factor $2$ out of $- 10 y$ .

$x' = \frac{x}{3 x - y} + \frac{2 (- 5 y)}{2 (3 x - y)}$

Step 5.3.3.1.2.2

Cancel the common factors.

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Step 5.3.3.1.2.2.1

Cancel the common factor.

$x' = \frac{x}{3 x - y} + \frac{2 (- 5 y)}{2 (3 x - y)}$

Step 5.3.3.1.2.2.2

Rewrite the expression.

$x' = \frac{x}{3 x - y} + \frac{- 5 y}{3 x - y}$

Step 5.3.3.1.3

Move the negative in front of the fraction.

$x' = \frac{x}{3 x - y} - \frac{5 y}{3 x - y}$

Step 5.3.3.2

Combine the numerators over the common denominator.

$x' = \frac{x - 5 y}{3 x - y}$

Step 6

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

$\frac{d x}{d y} = \frac{x - 5 y}{3 x - y}$