Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

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

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

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 $u_{1}$ as $x$ .

$3 (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d y} [x]) + \frac{d}{d y} [5 x^{2} y^{2}]$

Step 2.2.2.2

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

$3 (2 u_{1} \frac{d}{d y} [x]) + \frac{d}{d y} [5 x^{2} y^{2}]$

Step 2.2.2.3

Replace all occurrences of $u_{1}$ with $x$ .

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $3$ .

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

Step 2.3

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

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Step 2.3.1

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

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

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

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

Step 2.3.3

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

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

Step 2.3.4

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 2.3.4.1

To apply the Chain Rule, set $u_{2}$ as $x$ .

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

Step 2.3.4.2

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

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

Step 2.3.4.3

Replace all occurrences of $u_{2}$ with $x$ .

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

Step 2.3.5

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

$6 x x' + 5 (x^{2} (2 y) + y^{2} (2 x x'))$

Step 2.3.6

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

$6 x x' + 5 (2 \cdot x^{2} y + y^{2} (2 x x'))$

Step 2.3.7

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

$6 x x' + 5 (2 x^{2} y + 2 y^{2} x x')$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$6 x x' + 5 (2 x^{2} y) + 5 (2 y^{2} x x')$

Step 2.4.2

Combine terms.

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Step 2.4.2.1

Multiply $2$ by $5$ .

$6 x x' + 10 (x^{2} y) + 5 (2 y^{2} x x')$

Step 2.4.2.2

Multiply $2$ by $5$ .

$6 x x' + 10 x^{2} y + 10 y^{2} x x'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

$2 \frac{d}{d y} [y]$

Step 3.2

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

$2 \cdot 1$

Step 3.3

Multiply $2$ by $1$ .

$2$

Step 4

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

$6 x x' + 10 x^{2} y + 10 y^{2} x x' = 2$

Step 5

Solve for $x'$ .

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Step 5.1

Subtract $10 x^{2} y$ from both sides of the equation.

$6 x x' + 10 y^{2} x x' = 2 - 10 x^{2} y$

Step 5.2

Factor $2 x x'$ out of $6 x x' + 10 y^{2} x x'$ .

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Step 5.2.1

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

$2 x x' \cdot 3 + 10 y^{2} x x' = 2 - 10 x^{2} y$

Step 5.2.2

Factor $2 x x'$ out of $10 y^{2} x x'$ .

$2 x x' \cdot 3 + 2 x x' (5 y^{2}) = 2 - 10 x^{2} y$

Step 5.2.3

Factor $2 x x'$ out of $2 x x' \cdot 3 + 2 x x' (5 y^{2})$ .

$2 x x' (3 + 5 y^{2}) = 2 - 10 x^{2} y$

Step 5.3

Divide each term in $2 x x' (3 + 5 y^{2}) = 2 - 10 x^{2} y$ by $2 x (3 + 5 y^{2})$ and simplify.

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Step 5.3.1

Divide each term in $2 x x' (3 + 5 y^{2}) = 2 - 10 x^{2} y$ by $2 x (3 + 5 y^{2})$ .

$\frac{2 x x' (3 + 5 y^{2})}{2 x (3 + 5 y^{2})} = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

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.

$\frac{2 x x' (3 + 5 y^{2})}{2 x (3 + 5 y^{2})} = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

Step 5.3.2.1.2

Rewrite the expression.

$\frac{x x' (3 + 5 y^{2})}{x (3 + 5 y^{2})} = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

Step 5.3.2.2

Cancel the common factor of $x$ .

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Step 5.3.2.2.1

Cancel the common factor.

$\frac{x x' (3 + 5 y^{2})}{x (3 + 5 y^{2})} = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

Step 5.3.2.2.2

Rewrite the expression.

$\frac{x' (3 + 5 y^{2})}{3 + 5 y^{2}} = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

Step 5.3.2.3

Cancel the common factor of $3 + 5 y^{2}$ .

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Step 5.3.2.3.1

Cancel the common factor.

$\frac{x' (3 + 5 y^{2})}{3 + 5 y^{2}} = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

Step 5.3.2.3.2

Divide $x'$ by $1$ .

$x' = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

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.

$x' = \frac{2}{2 x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

Step 5.3.3.1.1.2

Rewrite the expression.

$x' = \frac{1}{x (3 + 5 y^{2})} + \frac{- 10 x^{2} y}{2 x (3 + 5 y^{2})}$

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 $- 10 x^{2} y$ .

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

Step 5.3.3.1.2.2

Cancel the common factors.

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Step 5.3.3.1.2.2.1

Factor $2$ out of $2 x (3 + 5 y^{2})$ .

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

Step 5.3.3.1.2.2.2

Cancel the common factor.

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

Step 5.3.3.1.2.2.3

Rewrite the expression.

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

Step 5.3.3.1.3

Cancel the common factor of $x^{2}$ and $x$ .

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Step 5.3.3.1.3.1

Factor $x$ out of $- 5 x^{2} y$ .

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

Step 5.3.3.1.3.2

Cancel the common factors.

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Step 5.3.3.1.3.2.1

Cancel the common factor.

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

Step 5.3.3.1.3.2.2

Rewrite the expression.

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

Step 5.3.3.1.4

Move the negative in front of the fraction.

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

Step 5.3.3.2

To write $- \frac{5 x y}{3 + 5 y^{2}}$ as a fraction with a common denominator, multiply by $\frac{x}{x}$ .

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

Step 5.3.3.3

Write each expression with a common denominator of $x (3 + 5 y^{2})$ , by multiplying each by an appropriate factor of $1$ .

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Step 5.3.3.3.1

Multiply $\frac{5 x y}{3 + 5 y^{2}}$ by $\frac{x}{x}$ .

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

Step 5.3.3.3.2

Reorder the factors of $(3 + 5 y^{2}) x$ .

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

Step 5.3.3.4

Combine the numerators over the common denominator.

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

Step 5.3.3.5

Multiply $x$ by $x$ by adding the exponents.

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Step 5.3.3.5.1

Move $x$ .

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

Step 5.3.3.5.2

Multiply $x$ by $x$ .

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

Step 6

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

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