Calculus Examples

$x^{5} + y^{5} = 40 x y$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{5} + y^{5}) = \frac{d}{d x} (40 x y)$

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 $x^{5} + y^{5}$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [y^{5}]$ .

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

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$5 x^{4} + \frac{d}{d x} [y^{5}]$

Step 2.2

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

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Step 2.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{5}$ and $g (x) = y$ .

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Step 2.2.1.1

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

$5 x^{4} + \frac{d}{d u} [u^{5}] \frac{d}{d x} [y]$

Step 2.2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d u} [u^{n}]$ is $n u^{n - 1}$ where $n = 5$ .

$5 x^{4} + 5 u^{4} \frac{d}{d x} [y]$

Step 2.2.1.3

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

$5 x^{4} + 5 y^{4} \frac{d}{d x} [y]$

Step 2.2.2

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

$5 x^{4} + 5 y^{4} y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $40$ is constant with respect to $x$ , the derivative of $40 x y$ with respect to $x$ is $40 \frac{d}{d x} [x y]$ .

$40 \frac{d}{d x} [x y]$

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 $g (x) = y$ .

$40 (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 3.3

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

$40 (x y' + y \frac{d}{d x} [x])$

Step 3.4

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

$40 (x y' + y \cdot 1)$

Step 3.5

Multiply $y$ by $1$ .

$40 (x y' + y)$

Step 3.6

Apply the distributive property.

$40 x y' + 40 y$

Step 4

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

$5 x^{4} + 5 y^{4} y' = 40 x y' + 40 y$

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $40 x y'$ from both sides of the equation.

$5 x^{4} + 5 y^{4} y' - 40 x y' = 40 y$

Step 5.2

Subtract $5 x^{4}$ from both sides of the equation.

$5 y^{4} y' - 40 x y' = 40 y - 5 x^{4}$

Step 5.3

Factor $5 y'$ out of $5 y^{4} y' - 40 x y'$ .

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Step 5.3.1

Factor $5 y'$ out of $5 y^{4} y'$ .

$5 y' y^{4} - 40 x y' = 40 y - 5 x^{4}$

Step 5.3.2

Factor $5 y'$ out of $- 40 x y'$ .

$5 y' y^{4} + 5 y' (- 8 x) = 40 y - 5 x^{4}$

Step 5.3.3

Factor $5 y'$ out of $5 y' y^{4} + 5 y' (- 8 x)$ .

$5 y' (y^{4} - 8 x) = 40 y - 5 x^{4}$

Step 5.4

Divide each term in $5 y' (y^{4} - 8 x) = 40 y - 5 x^{4}$ by $5 (y^{4} - 8 x)$ and simplify.

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Step 5.4.1

Divide each term in $5 y' (y^{4} - 8 x) = 40 y - 5 x^{4}$ by $5 (y^{4} - 8 x)$ .

$\frac{5 y' (y^{4} - 8 x)}{5 (y^{4} - 8 x)} = \frac{40 y}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $5$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{5 y' (y^{4} - 8 x)}{5 (y^{4} - 8 x)} = \frac{40 y}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (y^{4} - 8 x)}{y^{4} - 8 x} = \frac{40 y}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

Step 5.4.2.2

Cancel the common factor of $y^{4} - 8 x$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (y^{4} - 8 x)}{y^{4} - 8 x} = \frac{40 y}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{40 y}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

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 $40$ and $5$ .

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Step 5.4.3.1.1.1

Factor $5$ out of $40 y$ .

$y' = \frac{5 (8 y)}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

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.

$y' = \frac{5 (8 y)}{5 (y^{4} - 8 x)} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

Step 5.4.3.1.1.2.2

Rewrite the expression.

$y' = \frac{8 y}{y^{4} - 8 x} + \frac{- 5 x^{4}}{5 (y^{4} - 8 x)}$

Step 5.4.3.1.2

Cancel the common factor of $- 5$ and $5$ .

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Step 5.4.3.1.2.1

Factor $5$ out of $- 5 x^{4}$ .

$y' = \frac{8 y}{y^{4} - 8 x} + \frac{5 (- x^{4})}{5 (y^{4} - 8 x)}$

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.

$y' = \frac{8 y}{y^{4} - 8 x} + \frac{5 (- x^{4})}{5 (y^{4} - 8 x)}$

Step 5.4.3.1.2.2.2

Rewrite the expression.

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

Step 5.4.3.1.3

Move the negative in front of the fraction.

$y' = \frac{8 y}{y^{4} - 8 x} - \frac{x^{4}}{y^{4} - 8 x}$

Step 5.4.3.2

Combine the numerators over the common denominator.

$y' = \frac{8 y - x^{4}}{y^{4} - 8 x}$

Step 6

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

$\frac{d y}{d x} = \frac{8 y - x^{4}}{y^{4} - 8 x}$