Calculus Examples

$2 x^{2} + y^{2} - x y = 4$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (2 x^{2} + y^{2} - x y) = \frac{d}{d x} (4)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.2.3

Multiply $2$ by $2$ .

$4 x + \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- x y]$

Step 2.3

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

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Step 2.3.1

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

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Step 2.3.1.1

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

$4 x + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y] + \frac{d}{d x} [- x y]$

Step 2.3.1.2

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

$4 x + 2 u \frac{d}{d x} [y] + \frac{d}{d x} [- x y]$

Step 2.3.1.3

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

$4 x + 2 y \frac{d}{d x} [y] + \frac{d}{d x} [- x y]$

Step 2.3.2

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

$4 x + 2 y y' + \frac{d}{d x} [- x y]$

Step 2.4

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

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Step 2.4.1

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

$4 x + 2 y y' - \frac{d}{d x} [x y]$

Step 2.4.2

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 $g (x) = y$ .

$4 x + 2 y y' - (x \frac{d}{d x} [y] + y \frac{d}{d x} [x])$

Step 2.4.3

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

$4 x + 2 y y' - (x y' + y \frac{d}{d x} [x])$

Step 2.4.4

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

$4 x + 2 y y' - (x y' + y \cdot 1)$

Step 2.4.5

Multiply $y$ by $1$ .

$4 x + 2 y y' - (x y' + y)$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$4 x + 2 y y' - (x y') - y$

Step 2.5.2

Remove unnecessary parentheses.

$4 x + 2 y y' - x y' - y$

Step 2.5.3

Reorder terms.

$2 y y' - x y' + 4 x - y$

Step 3

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$0$

Step 4

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

$2 y y' - x y' + 4 x - y = 0$

Step 5

Solve for $y'$ .

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Step 5.1

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

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Step 5.1.1

Subtract $4 x$ from both sides of the equation.

$2 y y' - x y' - y = - 4 x$

Step 5.1.2

Add $y$ to both sides of the equation.

$2 y y' - x y' = - 4 x + y$

Step 5.2

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

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Step 5.2.1

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

$y' (2 y) - x y' = - 4 x + y$

Step 5.2.2

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

$y' (2 y) + y' (- x) = - 4 x + y$

Step 5.2.3

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

$y' (2 y - x) = - 4 x + y$

Step 5.3

Divide each term in $y' (2 y - x) = - 4 x + y$ by $2 y - x$ and simplify.

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Step 5.3.1

Divide each term in $y' (2 y - x) = - 4 x + y$ by $2 y - x$ .

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

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

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

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 4 x}{2 y - x} + \frac{y}{2 y - x}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 4 x + y}{2 y - x}$

Step 5.3.3.2

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

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

Step 5.3.3.3

Factor $- 1$ out of $y$ .

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

Step 5.3.3.4

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

$y' = \frac{- (4 x - y)}{2 y - x}$

Step 5.3.3.5

Simplify the expression.

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Step 5.3.3.5.1

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

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

Step 5.3.3.5.2

Move the negative in front of the fraction.

$y' = - \frac{4 x - y}{2 y - x}$

Step 6

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

$\frac{d y}{d x} = - \frac{4 x - y}{2 y - x}$