Calculus Examples

${(2 y + 1)}^{3} - 24 x = - 3$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} ({(2 y + 1)}^{3} - 24 x) = \frac{d}{d x} (- 3)$

Step 2

Differentiate the left side of the equation.

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Step 2.1

Use the Binomial Theorem.

$\frac{d}{d x} [{(2 y)}^{3} + 3 {(2 y)}^{2} \cdot 1 + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2

Differentiate using the Sum Rule.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Apply the product rule to $2 y$ .

$\frac{d}{d x} [2^{3} y^{3} + 3 {(2 y)}^{2} \cdot 1 + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.2

Raise $2$ to the power of $3$ .

$\frac{d}{d x} [8 y^{3} + 3 {(2 y)}^{2} \cdot 1 + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.3

Apply the product rule to $2 y$ .

$\frac{d}{d x} [8 y^{3} + 3 (2^{2} y^{2}) \cdot 1 + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.4

Raise $2$ to the power of $2$ .

$\frac{d}{d x} [8 y^{3} + 3 (4 y^{2}) \cdot 1 + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.5

Multiply $4$ by $3$ .

$\frac{d}{d x} [8 y^{3} + 12 y^{2} \cdot 1 + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.6

Multiply $12$ by $1$ .

$\frac{d}{d x} [8 y^{3} + 12 y^{2} + 3 (2 y) \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.7

Multiply $2$ by $3$ .

$\frac{d}{d x} [8 y^{3} + 12 y^{2} + 6 y \cdot 1^{2} + 1^{3} - 24 x]$

Step 2.2.1.8

One to any power is one.

$\frac{d}{d x} [8 y^{3} + 12 y^{2} + 6 y \cdot 1 + 1^{3} - 24 x]$

Step 2.2.1.9

Multiply $6$ by $1$ .

$\frac{d}{d x} [8 y^{3} + 12 y^{2} + 6 y + 1^{3} - 24 x]$

Step 2.2.1.10

One to any power is one.

$\frac{d}{d x} [8 y^{3} + 12 y^{2} + 6 y + 1 - 24 x]$

Step 2.2.2

By the Sum Rule, the derivative of $8 y^{3} + 12 y^{2} + 6 y + 1 - 24 x$ with respect to $x$ is $\frac{d}{d x} [8 y^{3}] + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$ .

$\frac{d}{d x} [8 y^{3}] + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.3

Evaluate $\frac{d}{d x} [8 y^{3}]$ .

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Step 2.3.1

Since $8$ is constant with respect to $x$ , the derivative of $8 y^{3}$ with respect to $x$ is $8 \frac{d}{d x} [y^{3}]$ .

$8 \frac{d}{d x} [y^{3}] + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

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

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Step 2.3.2.1

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

$8 (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d x} [y]) + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.3.2.2

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

$8 (3 {u_{1}}^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.3.2.3

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

$8 (3 y^{2} \frac{d}{d x} [y]) + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.3.3

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

$8 (3 y^{2} y') + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.3.4

Multiply $3$ by $8$ .

$24 y^{2} y' + \frac{d}{d x} [12 y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.4

Evaluate $\frac{d}{d x} [12 y^{2}]$ .

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Step 2.4.1

Since $12$ is constant with respect to $x$ , the derivative of $12 y^{2}$ with respect to $x$ is $12 \frac{d}{d x} [y^{2}]$ .

$24 y^{2} y' + 12 \frac{d}{d x} [y^{2}] + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

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

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Step 2.4.2.1

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

$24 y^{2} y' + 12 (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [y]) + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

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 $n = 2$ .

$24 y^{2} y' + 12 (2 u_{2} \frac{d}{d x} [y]) + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.4.2.3

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

$24 y^{2} y' + 12 (2 y \frac{d}{d x} [y]) + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.4.3

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

$24 y^{2} y' + 12 (2 y y') + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.4.4

Multiply $2$ by $12$ .

$24 y^{2} y' + 24 y y' + \frac{d}{d x} [6 y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.5

Evaluate $\frac{d}{d x} [6 y]$ .

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Step 2.5.1

Since $6$ is constant with respect to $x$ , the derivative of $6 y$ with respect to $x$ is $6 \frac{d}{d x} [y]$ .

$24 y^{2} y' + 24 y y' + 6 \frac{d}{d x} [y] + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.5.2

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

$24 y^{2} y' + 24 y y' + 6 y' + \frac{d}{d x} [1] + \frac{d}{d x} [- 24 x]$

Step 2.6

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$24 y^{2} y' + 24 y y' + 6 y' + 0 + \frac{d}{d x} [- 24 x]$

Step 2.7

Evaluate $\frac{d}{d x} [- 24 x]$ .

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Step 2.7.1

Since $- 24$ is constant with respect to $x$ , the derivative of $- 24 x$ with respect to $x$ is $- 24 \frac{d}{d x} [x]$ .

$24 y^{2} y' + 24 y y' + 6 y' + 0 - 24 \frac{d}{d x} [x]$

Step 2.7.2

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

$24 y^{2} y' + 24 y y' + 6 y' + 0 - 24 \cdot 1$

Step 2.7.3

Multiply $- 24$ by $1$ .

$24 y^{2} y' + 24 y y' + 6 y' + 0 - 24$

Step 2.8

Add $24 y^{2} y' + 24 y y' + 6 y'$ and $0$ .

$24 y^{2} y' + 24 y y' + 6 y' - 24$

Step 3

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

$0$

Step 4

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

$24 y^{2} y' + 24 y y' + 6 y' - 24 = 0$

Step 5

Solve for $y'$ .

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Step 5.1

Add $24$ to both sides of the equation.

$24 y^{2} y' + 24 y y' + 6 y' = 24$

Step 5.2

Factor $6 y'$ out of $24 y^{2} y' + 24 y y' + 6 y'$ .

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Step 5.2.1

Factor $6 y'$ out of $24 y^{2} y'$ .

$6 y' (4 y^{2}) + 24 y y' + 6 y' = 24$

Step 5.2.2

Factor $6 y'$ out of $24 y y'$ .

$6 y' (4 y^{2}) + 6 y' (4 y) + 6 y' = 24$

Step 5.2.3

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

$6 y' (4 y^{2}) + 6 y' (4 y) + 6 y' \cdot 1 = 24$

Step 5.2.4

Factor $6 y'$ out of $6 y' (4 y^{2}) + 6 y' (4 y)$ .

$6 y' (4 y^{2} + 4 y) + 6 y' \cdot 1 = 24$

Step 5.2.5

Factor $6 y'$ out of $6 y' (4 y^{2} + 4 y) + 6 y' \cdot 1$ .

$6 y' (4 y^{2} + 4 y + 1) = 24$

Step 5.3

Factor using the perfect square rule.

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Step 5.3.1

Rewrite $4 y^{2}$ as ${(2 y)}^{2}$ .

$6 y' ({(2 y)}^{2} + 4 y + 1) = 24$

Step 5.3.2

Rewrite $1$ as $1^{2}$ .

$6 y' ({(2 y)}^{2} + 4 y + 1^{2}) = 24$

Step 5.3.3

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 y = 2 \cdot (2 y) \cdot 1$

Step 5.3.4

Rewrite the polynomial.

$6 y' ({(2 y)}^{2} + 2 \cdot (2 y) \cdot 1 + 1^{2}) = 24$

Step 5.3.5

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 2 y$ and $b = 1$ .

$6 y' {(2 y + 1)}^{2} = 24$

Step 5.4

Divide each term in $6 y' {(2 y + 1)}^{2} = 24$ by $6 {(2 y + 1)}^{2}$ and simplify.

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Step 5.4.1

Divide each term in $6 y' {(2 y + 1)}^{2} = 24$ by $6 {(2 y + 1)}^{2}$ .

$\frac{6 y' {(2 y + 1)}^{2}}{6 {(2 y + 1)}^{2}} = \frac{24}{6 {(2 y + 1)}^{2}}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $6$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{6 y' {(2 y + 1)}^{2}}{6 {(2 y + 1)}^{2}} = \frac{24}{6 {(2 y + 1)}^{2}}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{y' {(2 y + 1)}^{2}}{{(2 y + 1)}^{2}} = \frac{24}{6 {(2 y + 1)}^{2}}$

Step 5.4.2.2

Cancel the common factor of ${(2 y + 1)}^{2}$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{y' {(2 y + 1)}^{2}}{{(2 y + 1)}^{2}} = \frac{24}{6 {(2 y + 1)}^{2}}$

Step 5.4.2.2.2

Divide $y'$ by $1$ .

$y' = \frac{24}{6 {(2 y + 1)}^{2}}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Cancel the common factor of $24$ and $6$ .

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Step 5.4.3.1.1

Factor $6$ out of $24$ .

$y' = \frac{6 \cdot 4}{6 {(2 y + 1)}^{2}}$

Step 5.4.3.1.2

Cancel the common factors.

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Step 5.4.3.1.2.1

Factor $6$ out of $6 {(2 y + 1)}^{2}$ .

$y' = \frac{6 \cdot 4}{6 ({(2 y + 1)}^{2})}$

Step 5.4.3.1.2.2

Cancel the common factor.

$y' = \frac{6 \cdot 4}{6 {(2 y + 1)}^{2}}$

Step 5.4.3.1.2.3

Rewrite the expression.

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

Step 6

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

$\frac{d y}{d x} = \frac{4}{{(2 y + 1)}^{2}}$