Calculus Examples

$- 1 = - x y + 2 y^{2}$

Step 1

Differentiate both sides of the equation.

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

Step 2

Since $- 1$ is constant with respect to $x$ , the derivative of $- 1$ with respect to $x$ is $0$ .

$0$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

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

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

Step 3.2.2

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 $g (x) = y$ .

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

Step 3.2.3

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

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

Step 3.2.4

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

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

Step 3.2.5

Multiply $y$ by $1$ .

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

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 3.3.2.1

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

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

Step 3.3.2.2

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

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

Step 3.3.2.3

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

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

Step 3.3.3

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

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

Step 3.3.4

Multiply $2$ by $2$ .

$- (x y' + y) + 4 y y'$

Step 3.4

Simplify.

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Step 3.4.1

Apply the distributive property.

$- (x y') - y + 4 y y'$

Step 3.4.2

Reorder terms.

$- x y' + 4 y y' - y$

Step 4

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

$0 = - x y' + 4 y y' - y$

Step 5

Solve for $y'$ .

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Step 5.1

Rewrite the equation as $- x y' + 4 y y' - y = 0$ .

$- x y' + 4 y y' - y = 0$

Step 5.2

Add $y$ to both sides of the equation.

$- x y' + 4 y y' = y$

Step 5.3

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

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Step 5.3.1

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

$y' (- x) + 4 y y' = y$

Step 5.3.2

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

$y' (- x) + y' (4 y) = y$

Step 5.3.3

Factor $y'$ out of $y' (- x) + y' (4 y)$ .

$y' (- x + 4 y) = y$

Step 5.4

Divide each term in $y' (- x + 4 y) = y$ by $- x + 4 y$ and simplify.

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Step 5.4.1

Divide each term in $y' (- x + 4 y) = y$ by $- x + 4 y$ .

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

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- x + 4 y$ .

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Step 5.4.2.1.1

Cancel the common factor.

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

Step 5.4.2.1.2

Divide $y'$ by $1$ .

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

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

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

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

Step 5.4.3.2

Factor $- 1$ out of $4 y$ .

$y' = \frac{y}{- (x) - (- 4 y)}$

Step 5.4.3.3

Factor $- 1$ out of $- (x) - (- 4 y)$ .

$y' = \frac{y}{- (x - 4 y)}$

Step 5.4.3.4

Rewrite negatives.

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Step 5.4.3.4.1

Rewrite $- (x - 4 y)$ as $- 1 (x - 4 y)$ .

$y' = \frac{y}{- 1 (x - 4 y)}$

Step 5.4.3.4.2

Move the negative in front of the fraction.

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

Step 6

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

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