Calculus Examples

$2 y^{3} - x^{2} + 2 y = - 4 y^{2} - x^{3}$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (2 y^{3} - x^{2} + 2 y) = \frac{d}{d x} (- 4 y^{2} - x^{3})$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $2 y^{3} - x^{2} + 2 y$ with respect to $x$ is $\frac{d}{d x} [2 y^{3}] + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 y]$ .

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

Step 2.2

Evaluate $\frac{d}{d x} [2 y^{3}]$ .

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Step 2.2.1

Since $2$ is constant with respect to $x$ , the derivative of $2 y^{3}$ with respect to $x$ is $2 \frac{d}{d x} [y^{3}]$ .

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

Step 2.2.2

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 2.2.2.1

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

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

Step 2.2.2.2

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

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

Step 2.2.2.3

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

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $3$ by $2$ .

$6 y^{2} y' + \frac{d}{d x} [- x^{2}] + \frac{d}{d x} [2 y]$

Step 2.3

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

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Step 2.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$6 y^{2} y' - \frac{d}{d x} [x^{2}] + \frac{d}{d x} [2 y]$

Step 2.3.2

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

$6 y^{2} y' - (2 x) + \frac{d}{d x} [2 y]$

Step 2.3.3

Multiply $2$ by $- 1$ .

$6 y^{2} y' - 2 x + \frac{d}{d x} [2 y]$

Step 2.4

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

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Step 2.4.1

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

$6 y^{2} y' - 2 x + 2 \frac{d}{d x} [y]$

Step 2.4.2

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

$6 y^{2} y' - 2 x + 2 y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $- 4 y^{2} - x^{3}$ with respect to $x$ is $\frac{d}{d x} [- 4 y^{2}] + \frac{d}{d x} [- x^{3}]$ .

$\frac{d}{d x} [- 4 y^{2}] + \frac{d}{d x} [- x^{3}]$

Step 3.2

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

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Step 3.2.1

Since $- 4$ is constant with respect to $x$ , the derivative of $- 4 y^{2}$ with respect to $x$ is $- 4 \frac{d}{d x} [y^{2}]$ .

$- 4 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- x^{3}]$

Step 3.2.2

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 3.2.2.1

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

$- 4 (\frac{d}{d u} [u^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [- x^{3}]$

Step 3.2.2.2

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

$- 4 (2 u \frac{d}{d x} [y]) + \frac{d}{d x} [- x^{3}]$

Step 3.2.2.3

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

$- 4 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [- x^{3}]$

Step 3.2.3

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

$- 4 (2 y y') + \frac{d}{d x} [- x^{3}]$

Step 3.2.4

Multiply $2$ by $- 4$ .

$- 8 y y' + \frac{d}{d x} [- x^{3}]$

Step 3.3

Evaluate $\frac{d}{d x} [- x^{3}]$ .

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Step 3.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{3}$ with respect to $x$ is $- \frac{d}{d x} [x^{3}]$ .

$- 8 y y' - \frac{d}{d x} [x^{3}]$

Step 3.3.2

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

$- 8 y y' - (3 x^{2})$

Step 3.3.3

Multiply $3$ by $- 1$ .

$- 8 y y' - 3 x^{2}$

Step 3.4

Reorder terms.

$- 3 x^{2} - 8 y y'$

Step 4

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

$6 y^{2} y' - 2 x + 2 y' = - 3 x^{2} - 8 y y'$

Step 5

Solve for $y'$ .

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Step 5.1

Add $8 y y'$ to both sides of the equation.

$6 y^{2} y' - 2 x + 2 y' + 8 y y' = - 3 x^{2}$

Step 5.2

Add $2 x$ to both sides of the equation.

$6 y^{2} y' + 2 y' + 8 y y' = - 3 x^{2} + 2 x$

Step 5.3

Factor $2 y'$ out of $6 y^{2} y' + 2 y' + 8 y y'$ .

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Step 5.3.1

Factor $2 y'$ out of $6 y^{2} y'$ .

$2 y' (3 y^{2}) + 2 y' + 8 y y' = - 3 x^{2} + 2 x$

Step 5.3.2

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

$2 y' (3 y^{2}) + 2 y' \cdot 1 + 8 y y' = - 3 x^{2} + 2 x$

Step 5.3.3

Factor $2 y'$ out of $8 y y'$ .

$2 y' (3 y^{2}) + 2 y' \cdot 1 + 2 y' (4 y) = - 3 x^{2} + 2 x$

Step 5.3.4

Factor $2 y'$ out of $2 y' (3 y^{2}) + 2 y' \cdot 1$ .

$2 y' (3 y^{2} + 1) + 2 y' (4 y) = - 3 x^{2} + 2 x$

Step 5.3.5

Factor $2 y'$ out of $2 y' (3 y^{2} + 1) + 2 y' (4 y)$ .

$2 y' (3 y^{2} + 1 + 4 y) = - 3 x^{2} + 2 x$

Step 5.4

Factor.

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Step 5.4.1

Factor by grouping.

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Step 5.4.1.1

Reorder terms.

$2 y' (3 y^{2} + 4 y + 1) = - 3 x^{2} + 2 x$

Step 5.4.1.2

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 3 \cdot 1 = 3$ and whose sum is $b = 4$ .

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Step 5.4.1.2.1

Factor $4$ out of $4 y$ .

$2 y' (3 y^{2} + 4 (y) + 1) = - 3 x^{2} + 2 x$

Step 5.4.1.2.2

Rewrite $4$ as $1$ plus $3$

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

Step 5.4.1.2.3

Apply the distributive property.

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

Step 5.4.1.2.4

Multiply $y$ by $1$ .

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

Step 5.4.1.3

Factor out the greatest common factor from each group.

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Step 5.4.1.3.1

Group the first two terms and the last two terms.

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

Step 5.4.1.3.2

Factor out the greatest common factor (GCF) from each group.

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

Step 5.4.1.4

Factor the polynomial by factoring out the greatest common factor, $3 y + 1$ .

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

Step 5.4.2

Remove unnecessary parentheses.

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

Step 5.5

Divide each term in $2 y' (3 y + 1) (y + 1) = - 3 x^{2} + 2 x$ by $2 (3 y + 1) (y + 1)$ and simplify.

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Step 5.5.1

Divide each term in $2 y' (3 y + 1) (y + 1) = - 3 x^{2} + 2 x$ by $2 (3 y + 1) (y + 1)$ .

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

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $2$ .

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Step 5.5.2.1.1

Cancel the common factor.

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

Step 5.5.2.1.2

Rewrite the expression.

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

Step 5.5.2.2

Cancel the common factor of $3 y + 1$ .

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Step 5.5.2.2.1

Cancel the common factor.

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

Step 5.5.2.2.2

Rewrite the expression.

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

Step 5.5.2.3

Cancel the common factor of $y + 1$ .

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Step 5.5.2.3.1

Cancel the common factor.

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

Step 5.5.2.3.2

Divide $y'$ by $1$ .

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

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Simplify each term.

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Step 5.5.3.1.1

Move the negative in front of the fraction.

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

Step 5.5.3.1.2

Cancel the common factor of $2$ .

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Step 5.5.3.1.2.1

Cancel the common factor.

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

Step 5.5.3.1.2.2

Rewrite the expression.

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

Step 6

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

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