Calculus Examples

$y = \arcsin (2 x + 1)$

Step 1

Differentiate both sides of the equation.

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

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

Differentiate using the chain rule, which states that $\frac{d}{d y} [f (g (y))]$ is $f' (g (y)) g' (y)$ where $f (y) = \arcsin (y)$ and $g (y) = 2 x + 1$ .

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Step 3.1.1

To apply the Chain Rule, set $u$ as $2 x + 1$ .

$\frac{d}{d u} [\arcsin (u)] \frac{d}{d y} [2 x + 1]$

Step 3.1.2

The derivative of $\arcsin (u)$ with respect to $u$ is $\frac{1}{\sqrt{1 - u^{2}}}$ .

$\frac{1}{\sqrt{1 - u^{2}}} \frac{d}{d y} [2 x + 1]$

Step 3.1.3

Replace all occurrences of $u$ with $2 x + 1$ .

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

Step 3.2

Differentiate.

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Step 3.2.1

By the Sum Rule, the derivative of $2 x + 1$ with respect to $y$ is $\frac{d}{d y} [2 x] + \frac{d}{d y} [1]$ .

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

Step 3.2.2

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

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

Step 3.3

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

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

Step 3.4

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

$\frac{1}{\sqrt{1 - {(2 x + 1)}^{2}}} (2 x' + 0)$

Step 3.5

Combine fractions.

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Step 3.5.1

Add $2 x'$ and $0$ .

$\frac{1}{\sqrt{1 - {(2 x + 1)}^{2}}} (2 x')$

Step 3.5.2

Combine $2$ and $\frac{1}{\sqrt{1 - {(2 x + 1)}^{2}}}$ .

$\frac{2}{\sqrt{1 - {(2 x + 1)}^{2}}} x'$

Step 3.5.3

Combine $\frac{2}{\sqrt{1 - {(2 x + 1)}^{2}}}$ and $x'$ .

$\frac{2 x'}{\sqrt{1 - {(2 x + 1)}^{2}}}$

Step 4

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

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

Step 5

Solve for $x'$ .

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Step 5.1

Rewrite the equation as $\frac{2 x'}{\sqrt{1 - {(2 x + 1)}^{2}}} = 1$ .

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

Step 5.2

Multiply both sides by $\sqrt{1 - {(2 x + 1)}^{2}}$ .

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

Step 5.3

Simplify.

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Step 5.3.1

Simplify the left side.

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Step 5.3.1.1

Cancel the common factor of $\sqrt{1 - {(2 x + 1)}^{2}}$ .

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Step 5.3.1.1.1

Cancel the common factor.

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

Step 5.3.1.1.2

Rewrite the expression.

$2 x' = 1 \sqrt{1 - {(2 x + 1)}^{2}}$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Simplify $1 \sqrt{1 - {(2 x + 1)}^{2}}$ .

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Step 5.3.2.1.1

Write the expression using exponents.

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Step 5.3.2.1.1.1

Multiply $\sqrt{1 - {(2 x + 1)}^{2}}$ by $1$ .

$2 x' = \sqrt{1 - {(2 x + 1)}^{2}}$

Step 5.3.2.1.1.2

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

$2 x' = \sqrt{1^{2} - {(2 x + 1)}^{2}}$

Step 5.3.2.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 2 x + 1$ .

$2 x' = \sqrt{(1 + 2 x + 1) (1 - (2 x + 1))}$

Step 5.3.2.1.3

Simplify.

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Step 5.3.2.1.3.1

Add $1$ and $1$ .

$2 x' = \sqrt{(2 x + 2) (1 - (2 x + 1))}$

Step 5.3.2.1.3.2

Factor $2$ out of $2 x + 2$ .

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Step 5.3.2.1.3.2.1

Factor $2$ out of $2 x$ .

$2 x' = \sqrt{(2 (x) + 2) (1 - (2 x + 1))}$

Step 5.3.2.1.3.2.2

Factor $2$ out of $2$ .

$2 x' = \sqrt{(2 (x) + 2 (1)) (1 - (2 x + 1))}$

Step 5.3.2.1.3.2.3

Factor $2$ out of $2 (x) + 2 (1)$ .

$2 x' = \sqrt{2 (x + 1) (1 - (2 x + 1))}$

Step 5.3.2.1.3.3

Apply the distributive property.

$2 x' = \sqrt{2 (x + 1) (1 - (2 x) - 1 \cdot 1)}$

Step 5.3.2.1.3.4

Multiply $2$ by $- 1$ .

$2 x' = \sqrt{2 (x + 1) (1 - 2 x - 1 \cdot 1)}$

Step 5.3.2.1.3.5

Multiply $- 1$ by $1$ .

$2 x' = \sqrt{2 (x + 1) (1 - 2 x - 1)}$

Step 5.3.2.1.3.6

Subtract $1$ from $1$ .

$2 x' = \sqrt{2 (x + 1) (- 2 x + 0)}$

Step 5.3.2.1.3.7

Add $- 2 x$ and $0$ .

$2 x' = \sqrt{2 (x + 1) \cdot - 2 x}$

Step 5.3.2.1.3.8

Multiply $- 2$ by $2$ .

$2 x' = \sqrt{- 4 (x + 1) x}$

Step 5.3.2.1.4

Rewrite $- 4 (x + 1) x$ as $2^{2} \cdot (- 1 (x + 1^{2}) x)$ .

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Step 5.3.2.1.4.1

Factor $4$ out of $- 4$ .

$2 x' = \sqrt{4 (- 1) (x + 1) x}$

Step 5.3.2.1.4.2

Rewrite $4$ as $2^{2}$ .

$2 x' = \sqrt{2^{2} \cdot - 1 (x + 1) x}$

Step 5.3.2.1.4.3

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

$2 x' = \sqrt{2^{2} \cdot - 1 (x + 1^{2}) x}$

Step 5.3.2.1.4.4

Add parentheses.

$2 x' = \sqrt{2^{2} \cdot - 1 ((x + 1^{2}) x)}$

Step 5.3.2.1.4.5

Add parentheses.

$2 x' = \sqrt{2^{2} \cdot (- 1 (x + 1^{2}) x)}$

Step 5.3.2.1.5

Pull terms out from under the radical.

$2 x' = 2 \sqrt{- 1 (x + 1^{2}) x}$

Step 5.3.2.1.6

Simplify the expression.

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Step 5.3.2.1.6.1

One to any power is one.

$2 x' = 2 \sqrt{- 1 (x + 1) x}$

Step 5.3.2.1.6.2

Reorder factors in $2 \sqrt{- 1 (x + 1) x}$ .

$2 x' = 2 \sqrt{- x (x + 1)}$

Step 5.4

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

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Step 5.4.1

Divide each term in $2 x' = 2 \sqrt{- x (x + 1)}$ by $2$ .

$\frac{2 x'}{2} = \frac{2 \sqrt{- x (x + 1)}}{2}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $2$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{2 x'}{2} = \frac{2 \sqrt{- x (x + 1)}}{2}$

Step 5.4.2.1.2

Divide $x'$ by $1$ .

$x' = \frac{2 \sqrt{- x (x + 1)}}{2}$

Step 5.4.3

Simplify the right side.

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Step 5.4.3.1

Cancel the common factor of $2$ .

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Step 5.4.3.1.1

Cancel the common factor.

$x' = \frac{2 \sqrt{- x (x + 1)}}{2}$

Step 5.4.3.1.2

Divide $\sqrt{- x (x + 1)}$ by $1$ .

$x' = \sqrt{- x (x + 1)}$

Step 6

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

$\frac{d x}{d y} = \sqrt{- x (x + 1)}$