Calculus Examples

$\arcsin (x) = \ln (y)$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d x} (\arcsin (x)) = \frac{d}{d x} (\ln (y))$

Step 2

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

$\frac{1}{\sqrt{1 - x^{2}}}$

Step 3

Differentiate the right side of the equation.

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Step 3.1

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

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Step 3.1.1

To apply the Chain Rule, set $u$ as $y$ .

$\frac{d}{d u} [\ln (u)] \frac{d}{d x} [y]$

Step 3.1.2

The derivative of $\ln (u)$ with respect to $u$ is $\frac{1}{u}$ .

$\frac{1}{u} \frac{d}{d x} [y]$

Step 3.1.3

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

$\frac{1}{y} \frac{d}{d x} [y]$

Step 3.2

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

$\frac{1}{y} y'$

Step 3.3

Combine $\frac{1}{y}$ and $y'$ .

$\frac{y'}{y}$

Step 4

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

$\frac{1}{\sqrt{1 - x^{2}}} = \frac{y'}{y}$

Step 5

Solve for $y'$ .

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Step 5.1

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

$\frac{y'}{y} = \frac{1}{\sqrt{1 - x^{2}}}$

Step 5.2

Multiply both sides by $y$ .

$\frac{y'}{y} y = \frac{1}{\sqrt{1 - x^{2}}} y$

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 $y$ .

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Step 5.3.1.1.1

Cancel the common factor.

$\frac{y'}{y} y = \frac{1}{\sqrt{1 - x^{2}}} y$

Step 5.3.1.1.2

Rewrite the expression.

$y' = \frac{1}{\sqrt{1 - x^{2}}} y$

Step 5.3.2

Simplify the right side.

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Step 5.3.2.1

Simplify $\frac{1}{\sqrt{1 - x^{2}}} y$ .

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Step 5.3.2.1.1

Simplify the denominator.

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Step 5.3.2.1.1.1

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

$y' = \frac{1}{\sqrt{1^{2} - x^{2}}} y$

Step 5.3.2.1.1.2

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 $b = x$ .

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

Step 5.3.2.1.2

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

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

Step 5.3.2.1.3

Combine and simplify the denominator.

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Step 5.3.2.1.3.1

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

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

Step 5.3.2.1.3.2

Raise $\sqrt{(1 + x) (1 - x)}$ to the power of $1$ .

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

Step 5.3.2.1.3.3

Raise $\sqrt{(1 + x) (1 - x)}$ to the power of $1$ .

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

Step 5.3.2.1.3.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.3.2.1.3.5

Add $1$ and $1$ .

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

Step 5.3.2.1.3.6

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

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Step 5.3.2.1.3.6.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{(1 + x) (1 - x)}$ as ${((1 + x) (1 - x))}^{\frac{1}{2}}$ .

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

Step 5.3.2.1.3.6.2

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 5.3.2.1.3.6.3

Combine $\frac{1}{2}$ and $2$ .

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

Step 5.3.2.1.3.6.4

Cancel the common factor of $2$ .

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Step 5.3.2.1.3.6.4.1

Cancel the common factor.

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

Step 5.3.2.1.3.6.4.2

Rewrite the expression.

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

Step 5.3.2.1.3.6.5

Simplify.

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

Step 5.3.2.1.4

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

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

Step 6

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

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