Calculus Examples

$y^{3} - 3 y^{2} - 9 y + 2 x^{2} + 20 x = - 27$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (y^{3} - 3 y^{2} - 9 y + 2 x^{2} + 20 x) = \frac{d}{d x} (- 27)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

By the Sum Rule, the derivative of $y^{3} - 3 y^{2} - 9 y + 2 x^{2} + 20 x$ with respect to $x$ is $\frac{d}{d x} [y^{3}] + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$ .

$\frac{d}{d x} [y^{3}] + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.2

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

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Step 2.2.1

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 2.2.1.1

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

$\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.2.1.2

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

$3 {u_{1}}^{2} \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.2.1.3

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

$3 y^{2} \frac{d}{d x} [y] + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.2.2

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

$3 y^{2} y' + \frac{d}{d x} [- 3 y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.3

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

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Step 2.3.1

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

$3 y^{2} y' - 3 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.3.2

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 2.3.2.1

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

$3 y^{2} y' - 3 (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.3.2.2

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

$3 y^{2} y' - 3 (2 u_{2} \frac{d}{d x} [y]) + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.3.2.3

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

$3 y^{2} y' - 3 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.3.3

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

$3 y^{2} y' - 3 (2 y y') + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.3.4

Multiply $2$ by $- 3$ .

$3 y^{2} y' - 6 y y' + \frac{d}{d x} [- 9 y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.4

Evaluate $\frac{d}{d x} [- 9 y]$ .

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Step 2.4.1

Since $- 9$ is constant with respect to $x$ , the derivative of $- 9 y$ with respect to $x$ is $- 9 \frac{d}{d x} [y]$ .

$3 y^{2} y' - 6 y y' - 9 \frac{d}{d x} [y] + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.4.2

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

$3 y^{2} y' - 6 y y' - 9 y' + \frac{d}{d x} [2 x^{2}] + \frac{d}{d x} [20 x]$

Step 2.5

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

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Step 2.5.1

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

$3 y^{2} y' - 6 y y' - 9 y' + 2 \frac{d}{d x} [x^{2}] + \frac{d}{d x} [20 x]$

Step 2.5.2

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

$3 y^{2} y' - 6 y y' - 9 y' + 2 (2 x) + \frac{d}{d x} [20 x]$

Step 2.5.3

Multiply $2$ by $2$ .

$3 y^{2} y' - 6 y y' - 9 y' + 4 x + \frac{d}{d x} [20 x]$

Step 2.6

Evaluate $\frac{d}{d x} [20 x]$ .

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Step 2.6.1

Since $20$ is constant with respect to $x$ , the derivative of $20 x$ with respect to $x$ is $20 \frac{d}{d x} [x]$ .

$3 y^{2} y' - 6 y y' - 9 y' + 4 x + 20 \frac{d}{d x} [x]$

Step 2.6.2

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

$3 y^{2} y' - 6 y y' - 9 y' + 4 x + 20 \cdot 1$

Step 2.6.3

Multiply $20$ by $1$ .

$3 y^{2} y' - 6 y y' - 9 y' + 4 x + 20$

Step 2.7

Reorder terms.

$3 y^{2} y' - 6 y y' + 4 x + 20 - 9 y'$

Step 3

Since $- 27$ is constant with respect to $x$ , the derivative of $- 27$ with respect to $x$ is $0$ .

$0$

Step 4

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

$3 y^{2} y' - 6 y y' + 4 x + 20 - 9 y' = 0$

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 $4 x$ from both sides of the equation.

$3 y^{2} y' - 6 y y' + 20 - 9 y' = - 4 x$

Step 5.1.2

Subtract $20$ from both sides of the equation.

$3 y^{2} y' - 6 y y' - 9 y' = - 4 x - 20$

Step 5.2

Factor $3 y'$ out of $3 y^{2} y' - 6 y y' - 9 y'$ .

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Step 5.2.1

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

$3 y' y^{2} - 6 y y' - 9 y' = - 4 x - 20$

Step 5.2.2

Factor $3 y'$ out of $- 6 y y'$ .

$3 y' y^{2} + 3 y' (- 2 y) - 9 y' = - 4 x - 20$

Step 5.2.3

Factor $3 y'$ out of $- 9 y'$ .

$3 y' y^{2} + 3 y' (- 2 y) + 3 y' \cdot - 3 = - 4 x - 20$

Step 5.2.4

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

$3 y' (y^{2} - 2 y) + 3 y' \cdot - 3 = - 4 x - 20$

Step 5.2.5

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

$3 y' (y^{2} - 2 y - 3) = - 4 x - 20$

Step 5.3

Factor.

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Step 5.3.1

Factor $y^{2} - 2 y - 3$ using the AC method.

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Step 5.3.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $- 2$ .

$- 3, 1$

Step 5.3.1.2

Write the factored form using these integers.

$3 y' ((y - 3) (y + 1)) = - 4 x - 20$

Step 5.3.2

Remove unnecessary parentheses.

$3 y' (y - 3) (y + 1) = - 4 x - 20$

Step 5.4

Divide each term in $3 y' (y - 3) (y + 1) = - 4 x - 20$ by $3 (y - 3) (y + 1)$ and simplify.

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Step 5.4.1

Divide each term in $3 y' (y - 3) (y + 1) = - 4 x - 20$ by $3 (y - 3) (y + 1)$ .

$\frac{3 y' (y - 3) (y + 1)}{3 (y - 3) (y + 1)} = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $3$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{3 y' (y - 3) (y + 1)}{3 (y - 3) (y + 1)} = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' (y - 3) (y + 1)}{(y - 3) (y + 1)} = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.2.2

Cancel the common factor of $y - 3$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' (y - 3) (y + 1)}{(y - 3) (y + 1)} = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.2.2.2

Rewrite the expression.

$\frac{y' (y + 1)}{y + 1} = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.2.3

Cancel the common factor of $y + 1$ .

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Step 5.4.2.3.1

Cancel the common factor.

$\frac{y' (y + 1)}{y + 1} = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.2.3.2

Divide $y'$ by $1$ .

$y' = \frac{- 4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Simplify each term.

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Step 5.4.3.1.1

Move the negative in front of the fraction.

$y' = - \frac{4 x}{3 (y - 3) (y + 1)} + \frac{- 20}{3 (y - 3) (y + 1)}$

Step 5.4.3.1.2

Move the negative in front of the fraction.

$y' = - \frac{4 x}{3 (y - 3) (y + 1)} - \frac{20}{3 (y - 3) (y + 1)}$

Step 6

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

$\frac{d y}{d x} = - \frac{4 x}{3 (y - 3) (y + 1)} - \frac{20}{3 (y - 3) (y + 1)}$