Calculus Examples

$x^{4} + 4 y = 8 x y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{4} + 4 y) = \frac{d}{d x} (8 x y)$

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 $x^{4} + 4 y$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [4 y]$ .

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

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 x y$ with respect to $x$ is $8 \frac{d}{d x} [x y]$ .

$8 \frac{d}{d x} [x y]$

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 $f (x) = x$ and $g (x) = y$ .

$8 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 3.3

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

$8 (x y' + y \frac{d}{d x} [x])$

Step 3.4

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

$8 (x y' + y \cdot 1)$

Step 3.5

Multiply $y$ by $1$ .

$8 (x y' + y)$

Step 3.6

Apply the distributive property.

$8 x y' + 8 y$

Step 4

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

$4 x^{3} + 4 y' = 8 x y' + 8 y$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $8 x y'$ from both sides of the equation.

$4 x^{3} + 4 y' - 8 x y' = 8 y$

Step 5.2

Subtract $4 x^{3}$ from both sides of the equation.

$4 y' - 8 x y' = 8 y - 4 x^{3}$

Step 5.3

Factor $4 y'$ out of $4 y' - 8 x y'$ .

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Step 5.3.1

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

$4 y' \cdot 1 - 8 x y' = 8 y - 4 x^{3}$

Step 5.3.2

Factor $4 y'$ out of $- 8 x y'$ .

$4 y' \cdot 1 + 4 y' (- 2 x) = 8 y - 4 x^{3}$

Step 5.3.3

Factor $4 y'$ out of $4 y' \cdot 1 + 4 y' (- 2 x)$ .

$4 y' (1 - 2 x) = 8 y - 4 x^{3}$

Step 5.4

Divide each term in $4 y' (1 - 2 x) = 8 y - 4 x^{3}$ by $4 (1 - 2 x)$ and simplify.

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Step 5.4.1

Divide each term in $4 y' (1 - 2 x) = 8 y - 4 x^{3}$ by $4 (1 - 2 x)$ .

$\frac{4 y' (1 - 2 x)}{4 (1 - 2 x)} = \frac{8 y}{4 (1 - 2 x)} + \frac{- 4 x^{3}}{4 (1 - 2 x)}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $4$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{4 y' (1 - 2 x)}{4 (1 - 2 x)} = \frac{8 y}{4 (1 - 2 x)} + \frac{- 4 x^{3}}{4 (1 - 2 x)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (1 - 2 x)}{1 - 2 x} = \frac{8 y}{4 (1 - 2 x)} + \frac{- 4 x^{3}}{4 (1 - 2 x)}$

Step 5.4.2.2

Cancel the common factor of $1 - 2 x$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (1 - 2 x)}{1 - 2 x} = \frac{8 y}{4 (1 - 2 x)} + \frac{- 4 x^{3}}{4 (1 - 2 x)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{8 y}{4 (1 - 2 x)} + \frac{- 4 x^{3}}{4 (1 - 2 x)}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Cancel the common factor of $8$ and $4$ .

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Step 5.4.3.1.1.1

Factor $4$ out of $8 y$ .

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

Step 5.4.3.1.1.2

Cancel the common factors.

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Step 5.4.3.1.1.2.1

Cancel the common factor.

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

Step 5.4.3.1.1.2.2

Rewrite the expression.

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

Step 5.4.3.1.2

Cancel the common factor of $- 4$ and $4$ .

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Step 5.4.3.1.2.1

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

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

Step 5.4.3.1.2.2

Cancel the common factors.

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Step 5.4.3.1.2.2.1

Cancel the common factor.

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

Step 5.4.3.1.2.2.2

Rewrite the expression.

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

Step 5.4.3.1.3

Move the negative in front of the fraction.

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

Step 5.4.3.2

Combine the numerators over the common denominator.

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

Step 6

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

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