Calculus Examples

$- y + 2 x^{2} - x = - 5 y^{3}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (- y + 2 x^{2} - x) = \frac{d}{d x} (- 5 y^{3})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

Step 2.2.2

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

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

Step 2.3

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

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Step 2.3.1

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

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

Step 2.3.2

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

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

Step 2.3.3

Multiply $2$ by $2$ .

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

Step 2.4

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

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Step 2.4.1

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

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

Step 2.4.2

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

$- y' + 4 x - 1 \cdot 1$

Step 2.4.3

Multiply $- 1$ by $1$ .

$- y' + 4 x - 1$

Step 2.5

Reorder terms.

$4 x - 1 - y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 y^{3}$ with respect to $x$ is $- 5 \frac{d}{d x} [y^{3}]$ .

$- 5 \frac{d}{d x} [y^{3}]$

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 $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.2.1

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

$- 5 (\frac{d}{d u} [u^{3}] \frac{d}{d x} [y])$

Step 3.2.2

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

$- 5 (3 u^{2} \frac{d}{d x} [y])$

Step 3.2.3

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

$- 5 (3 y^{2} \frac{d}{d x} [y])$

Step 3.3

Multiply $3$ by $- 5$ .

$- 15 (y^{2} \frac{d}{d x} [y])$

Step 3.4

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

$- 15 y^{2} y'$

Step 4

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

$4 x - 1 - y' = - 15 y^{2} y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $15 y^{2} y'$ to both sides of the equation.

$4 x - 1 - y' + 15 y^{2} y' = 0$

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 $4 x$ from both sides of the equation.

$- 1 - y' + 15 y^{2} y' = - 4 x$

Step 5.2.2

Add $1$ to both sides of the equation.

$- y' + 15 y^{2} y' = - 4 x + 1$

Step 5.3

Factor $y'$ out of $- y' + 15 y^{2} y'$ .

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Step 5.3.1

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

$y' \cdot - 1 + 15 y^{2} y' = - 4 x + 1$

Step 5.3.2

Factor $y'$ out of $15 y^{2} y'$ .

$y' \cdot - 1 + y' (15 y^{2}) = - 4 x + 1$

Step 5.3.3

Factor $y'$ out of $y' \cdot - 1 + y' (15 y^{2})$ .

$y' (- 1 + 15 y^{2}) = - 4 x + 1$

Step 5.4

Divide each term in $y' (- 1 + 15 y^{2}) = - 4 x + 1$ by $- 1 + 15 y^{2}$ and simplify.

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Step 5.4.1

Divide each term in $y' (- 1 + 15 y^{2}) = - 4 x + 1$ by $- 1 + 15 y^{2}$ .

$\frac{y' (- 1 + 15 y^{2})}{- 1 + 15 y^{2}} = \frac{- 4 x}{- 1 + 15 y^{2}} + \frac{1}{- 1 + 15 y^{2}}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $- 1 + 15 y^{2}$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{y' (- 1 + 15 y^{2})}{- 1 + 15 y^{2}} = \frac{- 4 x}{- 1 + 15 y^{2}} + \frac{1}{- 1 + 15 y^{2}}$

Step 5.4.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 4 x}{- 1 + 15 y^{2}} + \frac{1}{- 1 + 15 y^{2}}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Combine the numerators over the common denominator.

$y' = \frac{- 4 x + 1}{- 1 + 15 y^{2}}$

Step 5.4.3.2

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

$y' = \frac{- (4 x) + 1}{- 1 + 15 y^{2}}$

Step 5.4.3.3

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

$y' = \frac{- (4 x) - 1 (- 1)}{- 1 + 15 y^{2}}$

Step 5.4.3.4

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

$y' = \frac{- (4 x - 1)}{- 1 + 15 y^{2}}$

Step 5.4.3.5

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

$y' = \frac{- 1 (4 x - 1)}{- 1 + 15 y^{2}}$

Step 5.4.3.6

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

$y' = \frac{- 1 (4 x - 1)}{- 1 (1) + 15 y^{2}}$

Step 5.4.3.7

Factor $- 1$ out of $15 y^{2}$ .

$y' = \frac{- 1 (4 x - 1)}{- 1 (1) - (- 15 y^{2})}$

Step 5.4.3.8

Factor $- 1$ out of $- 1 (1) - (- 15 y^{2})$ .

$y' = \frac{- 1 (4 x - 1)}{- 1 (1 - 15 y^{2})}$

Step 5.4.3.9

Cancel the common factor.

$y' = \frac{- 1 (4 x - 1)}{- 1 (1 - 15 y^{2})}$

Step 5.4.3.10

Rewrite the expression.

$y' = \frac{4 x - 1}{1 - 15 y^{2}}$

Step 6

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

$\frac{d y}{d x} = \frac{4 x - 1}{1 - 15 y^{2}}$