Calculus Examples

$- x y + 2 y + 5 x^{2} = - 4$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (- x y + 2 y + 5 x^{2}) = \frac{d}{d x} (- 4)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

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

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

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 $g (x) = y$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Multiply $y$ by $1$ .

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.4

Evaluate $\frac{d}{d x} [5 x^{2}]$ .

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Step 2.4.1

Since $5$ is constant with respect to $x$ , the derivative of $5 x^{2}$ with respect to $x$ is $5 \frac{d}{d x} [x^{2}]$ .

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

Step 2.4.2

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

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

Step 2.4.3

Multiply $2$ by $5$ .

$- (x y' + y) + 2 y' + 10 x$

Step 2.5

Simplify.

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Step 2.5.1

Apply the distributive property.

$- (x y') - y + 2 y' + 10 x$

Step 2.5.2

Reorder terms.

$- x y' + 10 x - y + 2 y'$

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.

$- x y' + 10 x - y + 2 y' = 0$

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 $10 x$ from both sides of the equation.

$- x y' - y + 2 y' = - 10 x$

Step 5.1.2

Add $y$ to both sides of the equation.

$- x y' + 2 y' = - 10 x + y$

Step 5.2

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

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Step 5.2.1

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

$y' (- x) + 2 y' = - 10 x + y$

Step 5.2.2

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

$y' (- x) + y' \cdot 2 = - 10 x + y$

Step 5.2.3

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

$y' (- x + 2) = - 10 x + y$

Step 5.3

Divide each term in $y' (- x + 2) = - 10 x + y$ by $- x + 2$ and simplify.

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Step 5.3.1

Divide each term in $y' (- x + 2) = - 10 x + y$ by $- x + 2$ .

$\frac{y' (- x + 2)}{- x + 2} = \frac{- 10 x}{- x + 2} + \frac{y}{- x + 2}$

Step 5.3.2

Simplify the left side.

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Step 5.3.2.1

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

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Step 5.3.2.1.1

Cancel the common factor.

$\frac{y' (- x + 2)}{- x + 2} = \frac{- 10 x}{- x + 2} + \frac{y}{- x + 2}$

Step 5.3.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 10 x}{- x + 2} + \frac{y}{- x + 2}$

Step 5.3.3

Simplify the right side.

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Step 5.3.3.1

Move the negative in front of the fraction.

$y' = - \frac{10 x}{- x + 2} + \frac{y}{- x + 2}$

Step 5.3.3.2

Combine the numerators over the common denominator.

$y' = \frac{- 10 x + y}{- x + 2}$

Step 5.3.3.3

Factor $- 1$ out of $- 10 x$ .

$y' = \frac{- (10 x) + y}{- x + 2}$

Step 5.3.3.4

Factor $- 1$ out of $y$ .

$y' = \frac{- (10 x) - 1 (- y)}{- x + 2}$

Step 5.3.3.5

Factor $- 1$ out of $- (10 x) - 1 (- y)$ .

$y' = \frac{- (10 x - y)}{- x + 2}$

Step 5.3.3.6

Rewrite $- (10 x - y)$ as $- 1 (10 x - y)$ .

$y' = \frac{- 1 (10 x - y)}{- x + 2}$

Step 5.3.3.7

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

$y' = \frac{- 1 (10 x - y)}{- (x) + 2}$

Step 5.3.3.8

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

$y' = \frac{- 1 (10 x - y)}{- (x) - 1 (- 2)}$

Step 5.3.3.9

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

$y' = \frac{- 1 (10 x - y)}{- (x - 2)}$

Step 5.3.3.10

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

$y' = \frac{- 1 (10 x - y)}{- 1 (x - 2)}$

Step 5.3.3.11

Cancel the common factor.

$y' = \frac{- 1 (10 x - y)}{- 1 (x - 2)}$

Step 5.3.3.12

Rewrite the expression.

$y' = \frac{10 x - y}{x - 2}$

Step 6

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

$\frac{d y}{d x} = \frac{10 x - y}{x - 2}$