Calculus Examples

$2 x + 5 y = x y$

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

$2 \cdot 1 + \frac{d}{d x} [5 y]$

Step 2.2.3

Multiply $2$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

$2 + 5 y'$

Step 2.4

Reorder terms.

$5 y' + 2$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

$x y' + y \cdot 1$

Step 3.4

Multiply $y$ by $1$ .

$x y' + y$

Step 4

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

$5 y' + 2 = x y' + y$

Step 5

Solve for $y'$ .

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Step 5.1

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

$5 y' + 2 - x y' = y$

Step 5.2

Subtract $2$ from both sides of the equation.

$5 y' - x y' = y - 2$

Step 5.3

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

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Step 5.3.1

Factor $y'$ out of $5 y'$ .

$y' \cdot 5 - x y' = y - 2$

Step 5.3.2

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

$y' \cdot 5 + y' (- x) = y - 2$

Step 5.3.3

Factor $y'$ out of $y' \cdot 5 + y' (- x)$ .

$y' (5 - x) = y - 2$

Step 5.4

Divide each term in $y' (5 - x) = y - 2$ by $5 - x$ and simplify.

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Step 5.4.1

Divide each term in $y' (5 - x) = y - 2$ by $5 - x$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $5 - x$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

$y' = \frac{y - 2}{5 - x}$

Step 6

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

$\frac{d y}{d x} = \frac{y - 2}{5 - x}$