Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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 $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

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

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

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

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

Step 2.4.2

Reorder terms.

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

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- y^{2}$ with respect to $x$ is $- \frac{d}{d x} [y^{2}]$ .

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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

$- (2 u \frac{d}{d x} [y])$

Step 3.2.3

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

$- (2 y \frac{d}{d x} [y])$

Step 3.3

Multiply $2$ by $- 1$ .

$- 2 (y \frac{d}{d x} [y])$

Step 3.4

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

$- 2 y y'$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Add $2 y y'$ to both sides of the equation.

$5 x y' + 2 x + 5 y + 2 y y' = 0$

Step 5.2

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

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Step 5.2.1

Subtract $2 x$ from both sides of the equation.

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

Step 5.2.2

Subtract $5 y$ from both sides of the equation.

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

Step 5.3

Factor $y'$ out of $5 x y' + 2 y y'$ .

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Step 5.3.1

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

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

Step 5.3.2

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

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

Step 5.3.3

Factor $y'$ out of $y' (5 x) + y' (2 y)$ .

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

Step 5.4

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

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Step 5.4.1

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

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $5 x + 2 y$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

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

Step 5.4.3.2

Factor $- 1$ out of $- 2 x$ .

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

Step 5.4.3.3

Factor $- 1$ out of $- 5 y$ .

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

Step 5.4.3.4

Factor $- 1$ out of $- (2 x) - (5 y)$ .

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

Step 5.4.3.5

Simplify the expression.

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Step 5.4.3.5.1

Rewrite $- (2 x + 5 y)$ as $- 1 (2 x + 5 y)$ .

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

Step 5.4.3.5.2

Move the negative in front of the fraction.

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

Step 6

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

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