Calculus Examples

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

Step 1

Differentiate both sides of the equation.

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

Step 2

Differentiate the left side of the equation.

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Step 2.1

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

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

Step 2.2

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

$- y'$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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

Step 3.2

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

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Step 3.2.1

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{3}$ and $g (x) = y$ .

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Step 3.2.1.1

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

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

Step 3.2.1.2

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

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

Step 3.2.1.3

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

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

Step 3.2.2

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

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

Step 3.3

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

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Step 3.3.1

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

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

Step 3.3.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{2}$ and $g (x) = y$ .

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Step 3.3.2.1

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

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

Step 3.3.2.2

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

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

Step 3.3.2.3

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

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

Step 3.3.3

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

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

Step 3.3.4

Multiply $2$ by $- 1$ .

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

Step 3.4

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

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Step 3.4.1

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

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

Step 3.4.2

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

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

Step 3.4.3

Multiply $3$ by $3$ .

$3 y^{2} y' - 2 y y' + 9 x^{2}$

Step 3.5

Reorder terms.

$9 x^{2} + 3 y^{2} y' - 2 y y'$

Step 4

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

$- y' = 9 x^{2} + 3 y^{2} y' - 2 y y'$

Step 5

Solve for $y'$ .

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Step 5.1

Since $y'$ is on the right side of the equation, switch the sides so it is on the left side of the equation.

$9 x^{2} + 3 y^{2} y' - 2 y y' = - y'$

Step 5.2

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

$9 x^{2} + 3 y^{2} y' - 2 y y' + y' = 0$

Step 5.3

Subtract $9 x^{2}$ from both sides of the equation.

$3 y^{2} y' - 2 y y' + y' = - 9 x^{2}$

Step 5.4

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

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Step 5.4.1

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

$y' (3 y^{2}) - 2 y y' + y' = - 9 x^{2}$

Step 5.4.2

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

$y' (3 y^{2}) + y' (- 2 y) + y' = - 9 x^{2}$

Step 5.4.3

Raise $y'$ to the power of $1$ .

$y' (3 y^{2}) + y' (- 2 y) + y' = - 9 x^{2}$

Step 5.4.4

Factor $y'$ out of ${y'}^{1}$ .

$y' (3 y^{2}) + y' (- 2 y) + y' \cdot 1 = - 9 x^{2}$

Step 5.4.5

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

$y' (3 y^{2} - 2 y) + y' \cdot 1 = - 9 x^{2}$

Step 5.4.6

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

$y' (3 y^{2} - 2 y + 1) = - 9 x^{2}$

Step 5.5

Divide each term in $y' (3 y^{2} - 2 y + 1) = - 9 x^{2}$ by $3 y^{2} - 2 y + 1$ and simplify.

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Step 5.5.1

Divide each term in $y' (3 y^{2} - 2 y + 1) = - 9 x^{2}$ by $3 y^{2} - 2 y + 1$ .

$\frac{y' (3 y^{2} - 2 y + 1)}{3 y^{2} - 2 y + 1} = \frac{- 9 x^{2}}{3 y^{2} - 2 y + 1}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $3 y^{2} - 2 y + 1$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{y' (3 y^{2} - 2 y + 1)}{3 y^{2} - 2 y + 1} = \frac{- 9 x^{2}}{3 y^{2} - 2 y + 1}$

Step 5.5.2.1.2

Divide $y'$ by $1$ .

$y' = \frac{- 9 x^{2}}{3 y^{2} - 2 y + 1}$

Step 5.5.3

Simplify the right side.

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Step 5.5.3.1

Move the negative in front of the fraction.

$y' = - \frac{9 x^{2}}{3 y^{2} - 2 y + 1}$

Step 6

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

$\frac{d y}{d x} = - \frac{9 x^{2}}{3 y^{2} - 2 y + 1}$