Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

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Step 2.2.1

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

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

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 $f (x) = x^{2}$ and $g (x) = y$ .

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Step 2.2.2.1

To apply the Chain Rule, set $u_{1}$ as $y$ .

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

Step 2.2.2.2

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

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

Step 2.2.2.3

Replace all occurrences of $u_{1}$ with $y$ .

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

Step 2.2.3

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

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

Step 2.2.4

Multiply $2$ by $2$ .

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

Step 2.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

$4 y y' + 0 + \frac{d}{d x} [- y^{3}] + \frac{d}{d x} [x^{3}]$

Step 2.4

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

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Step 2.4.1

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

$4 y y' + 0 - \frac{d}{d x} [y^{3}] + \frac{d}{d x} [x^{3}]$

Step 2.4.2

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 2.4.2.1

To apply the Chain Rule, set $u_{2}$ as $y$ .

$4 y y' + 0 - (\frac{d}{d u_{2}} [{u_{2}}^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [x^{3}]$

Step 2.4.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{2}} [{u_{2}}^{n}]$ is $n {u_{2}}^{n - 1}$ where $n = 3$ .

$4 y y' + 0 - (3 {u_{2}}^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x^{3}]$

Step 2.4.2.3

Replace all occurrences of $u_{2}$ with $y$ .

$4 y y' + 0 - (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [x^{3}]$

Step 2.4.3

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

$4 y y' + 0 - (3 y^{2} y') + \frac{d}{d x} [x^{3}]$

Step 2.4.4

Multiply $3$ by $- 1$ .

$4 y y' + 0 - 3 y^{2} y' + \frac{d}{d x} [x^{3}]$

Step 2.5

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

$4 y y' + 0 - 3 y^{2} y' + 3 x^{2}$

Step 2.6

Simplify.

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Step 2.6.1

Add $4 y y'$ and $0$ .

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

Step 2.6.2

Reorder terms.

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

Step 3

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

$y'$

Step 4

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

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

Step 5

Solve for $y'$ .

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Step 5.1

Subtract $y'$ from both sides of the equation.

$3 x^{2} + 4 y y' - 3 y^{2} y' - y' = 0$

Step 5.2

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

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

Step 5.3

Factor $y'$ out of $4 y y' - 3 y^{2} y' - y'$ .

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Step 5.3.1

Factor $y'$ out of $4 y y'$ .

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

Step 5.3.2

Factor $y'$ out of $- 3 y^{2} y'$ .

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

Step 5.3.3

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

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

Step 5.3.4

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

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

Step 5.3.5

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

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

Step 5.4

Let $u = y$ . Substitute $u$ for all occurrences of $y$ .

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

Step 5.5

Factor by grouping.

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Step 5.5.1

Reorder terms.

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

Step 5.5.2

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 5.5.2.1

Factor $4$ out of $4 u$ .

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

Step 5.5.2.2

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

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

Step 5.5.2.3

Apply the distributive property.

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

Step 5.5.2.4

Multiply $u$ by $1$ .

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

Step 5.5.3

Factor out the greatest common factor from each group.

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Step 5.5.3.1

Group the first two terms and the last two terms.

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

Step 5.5.3.2

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

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

Step 5.5.4

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

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

Step 5.6

Factor.

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Step 5.6.1

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

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

Step 5.6.2

Remove unnecessary parentheses.

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

Step 5.7

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

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Step 5.7.1

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

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

Step 5.7.2

Simplify the left side.

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Step 5.7.2.1

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

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Step 5.7.2.1.1

Cancel the common factor.

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

Step 5.7.2.1.2

Rewrite the expression.

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

Step 5.7.2.2

Cancel the common factor of $y - 1$ .

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Step 5.7.2.2.1

Cancel the common factor.

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

Step 5.7.2.2.2

Divide $y'$ by $1$ .

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

Step 5.7.3

Simplify the right side.

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Step 5.7.3.1

Move the negative in front of the fraction.

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

Step 5.7.3.2

Factor $- 1$ out of $- 3 y$ .

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

Step 5.7.3.3

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

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

Step 5.7.3.4

Factor $- 1$ out of $- (3 y) - 1 (- 1)$ .

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

Step 5.7.3.5

Simplify the expression.

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Step 5.7.3.5.1

Rewrite $- (3 y - 1)$ as $- 1 (3 y - 1)$ .

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

Step 5.7.3.5.2

Move the negative in front of the fraction.

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

Step 5.7.3.5.3

Multiply $- 1$ by $- 1$ .

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

Step 5.7.3.5.4

Multiply $\frac{3 x^{2}}{(3 y - 1) (y - 1)}$ by $1$ .

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

Step 6

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

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