Calculus Examples

$x^{3} - x y + y^{2} = 7$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (x^{3} - x y + y^{2}) = \frac{d}{d x} (7)$

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 $x^{3} - x y + y^{2}$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x y] + \frac{d}{d x} [y^{2}]$ .

$\frac{d}{d x} [x^{3}] + \frac{d}{d x} [- x y] + \frac{d}{d x} [y^{2}]$

Step 2.1.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

$3 x^{2} + \frac{d}{d x} [- x y] + \frac{d}{d x} [y^{2}]$

Step 2.2

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

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Step 2.2.1

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

$3 x^{2} - \frac{d}{d x} [x y] + \frac{d}{d x} [y^{2}]$

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 $g (x) = y$ .

$3 x^{2} - (x \frac{d}{d x} [y] + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}]$

Step 2.2.3

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

$3 x^{2} - (x y' + y \frac{d}{d x} [x]) + \frac{d}{d x} [y^{2}]$

Step 2.2.4

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

$3 x^{2} - (x y' + y \cdot 1) + \frac{d}{d x} [y^{2}]$

Step 2.2.5

Multiply $y$ by $1$ .

$3 x^{2} - (x y' + y) + \frac{d}{d x} [y^{2}]$

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 $y$ .

$3 x^{2} - (x y' + y) + \frac{d}{d u} [u^{2}] \frac{d}{d x} [y]$

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 $n = 2$ .

$3 x^{2} - (x y' + y) + 2 u \frac{d}{d x} [y]$

Step 2.3.1.3

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

$3 x^{2} - (x y' + y) + 2 y \frac{d}{d x} [y]$

Step 2.3.2

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

$3 x^{2} - (x y' + y) + 2 y y'$

Step 2.4

Simplify.

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Step 2.4.1

Apply the distributive property.

$3 x^{2} - (x y') - y + 2 y y'$

Step 2.4.2

Remove unnecessary parentheses.

$3 x^{2} - x y' - y + 2 y y'$

Step 2.4.3

Reorder terms.

$3 x^{2} - x y' + 2 y y' - y$

Step 3

Since $7$ is constant with respect to $x$ , the derivative of $7$ with respect to $x$ is $0$ .

$0$

Step 4

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

$3 x^{2} - x y' + 2 y y' - y = 0$

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 $3 x^{2}$ from both sides of the equation.

$- x y' + 2 y y' - y = - 3 x^{2}$

Step 5.1.2

Add $y$ to both sides of the equation.

$- x y' + 2 y y' = - 3 x^{2} + y$

Step 5.2

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

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Step 5.2.1

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

$y' (- x) + 2 y y' = - 3 x^{2} + y$

Step 5.2.2

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

$y' (- x) + y' (2 y) = - 3 x^{2} + y$

Step 5.2.3

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

$y' (- x + 2 y) = - 3 x^{2} + y$

Step 5.3

Divide each term in $y' (- x + 2 y) = - 3 x^{2} + y$ by $- x + 2 y$ and simplify.

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Step 5.3.1

Divide each term in $y' (- x + 2 y) = - 3 x^{2} + y$ by $- x + 2 y$ .

$\frac{y' (- x + 2 y)}{- x + 2 y} = \frac{- 3 x^{2}}{- x + 2 y} + \frac{y}{- x + 2 y}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (- x + 2 y)}{- x + 2 y} = \frac{- 3 x^{2}}{- x + 2 y} + \frac{y}{- x + 2 y}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 3 x^{2}}{- x + 2 y} + \frac{y}{- x + 2 y}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 3 x^{2} + y}{- x + 2 y}$

Step 5.3.3.2

Factor $- 1$ out of $- 3 x^{2}$ .

$y' = \frac{- (3 x^{2}) + y}{- x + 2 y}$

Step 5.3.3.3

Factor $- 1$ out of $y$ .

$y' = \frac{- (3 x^{2}) - 1 (- y)}{- x + 2 y}$

Step 5.3.3.4

Factor $- 1$ out of $- (3 x^{2}) - 1 (- y)$ .

$y' = \frac{- (3 x^{2} - y)}{- x + 2 y}$

Step 5.3.3.5

Rewrite $- (3 x^{2} - y)$ as $- 1 (3 x^{2} - y)$ .

$y' = \frac{- 1 (3 x^{2} - y)}{- x + 2 y}$

Step 5.3.3.6

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

$y' = \frac{- 1 (3 x^{2} - y)}{- (x) + 2 y}$

Step 5.3.3.7

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

$y' = \frac{- 1 (3 x^{2} - y)}{- (x) - (- 2 y)}$

Step 5.3.3.8

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

$y' = \frac{- 1 (3 x^{2} - y)}{- (x - 2 y)}$

Step 5.3.3.9

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

$y' = \frac{- 1 (3 x^{2} - y)}{- 1 (x - 2 y)}$

Step 5.3.3.10

Cancel the common factor.

$y' = \frac{- 1 (3 x^{2} - y)}{- 1 (x - 2 y)}$

Step 5.3.3.11

Rewrite the expression.

$y' = \frac{3 x^{2} - y}{x - 2 y}$

Step 6

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

$\frac{d y}{d x} = \frac{3 x^{2} - y}{x - 2 y}$