Calculus Examples

$h (x) = \arctan (x^{2})$

Find the first derivative.

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

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To apply the Chain Rule, set $u$ as $x^{2}$ .

$f' (x) = \frac{d}{d u} (\arctan (u)) \frac{d}{d x} (x^{2})$

The derivative of $\arctan (u)$ with respect to $u$ is $\frac{1}{1 + u^{2}}$ .

$f' (x) = \frac{1}{1 + u^{2}} \cdot \frac{d}{d x} (x^{2})$

Replace all occurrences of $u$ with $x^{2}$ .

$f' (x) = \frac{1}{1 + {(x^{2})}^{2}} \cdot \frac{d}{d x} (x^{2})$

Differentiate using the Power Rule.

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Multiply the exponents in ${(x^{2})}^{2}$ .

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

$f' (x) = \frac{1}{1 + x^{2 \cdot 2}} \cdot \frac{d}{d x} (x^{2})$

Multiply $2$ by $2$ .

$f' (x) = \frac{1}{1 + x^{4}} \cdot \frac{d}{d x} (x^{2})$

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

$f' (x) = \frac{1}{1 + x^{4}} \cdot (2 x)$

Combine fractions.

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Combine $2$ and $\frac{1}{1 + x^{4}}$ .

$f' (x) = \frac{2}{1 + x^{4}} \cdot x$

Combine $\frac{2}{1 + x^{4}}$ and $x$ .

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

Reorder terms.

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

Find the second derivative.

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Since $2$ is constant with respect to $x$ , the derivative of $\frac{2 x}{x^{4} + 1}$ with respect to $x$ is $2 \frac{d}{d x} [\frac{x}{x^{4} + 1}]$ .

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

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x^{4} + 1$ .

$f'' (x) = 2 (\frac{(x^{4} + 1) \frac{d}{d x} (x) - x \frac{d}{d x} (x^{4} + 1)}{{(x^{4} + 1)}^{2}})$

Differentiate.

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

$f'' (x) = 2 (\frac{(x^{4} + 1) \cdot 1 - x \frac{d}{d x} (x^{4} + 1)}{{(x^{4} + 1)}^{2}})$

Multiply $x^{4} + 1$ by $1$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - x \frac{d}{d x} (x^{4} + 1)}{{(x^{4} + 1)}^{2}})$

By the Sum Rule, the derivative of $x^{4} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [1]$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - x (\frac{d}{d x} (x^{4}) + \frac{d}{d x} (1))}{{(x^{4} + 1)}^{2}})$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - x (4 x^{3} + \frac{d}{d x} (1))}{{(x^{4} + 1)}^{2}})$

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

$f'' (x) = 2 (\frac{x^{4} + 1 - x (4 x^{3} + 0)}{{(x^{4} + 1)}^{2}})$

Simplify the expression.

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Add $4 x^{3}$ and $0$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - x (4 x^{3})}{{(x^{4} + 1)}^{2}})$

Multiply $4$ by $- 1$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - 4 x \cdot x^{3}}{{(x^{4} + 1)}^{2}})$

Multiply $x$ by $x^{3}$ by adding the exponents.

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Move $x^{3}$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - 4 (x^{3} x)}{{(x^{4} + 1)}^{2}})$

Multiply $x^{3}$ by $x$ .

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Raise $x$ to the power of $1$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - 4 (x^{3} x)}{{(x^{4} + 1)}^{2}})$

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

$f'' (x) = 2 (\frac{x^{4} + 1 - 4 x^{3 + 1}}{{(x^{4} + 1)}^{2}})$

Add $3$ and $1$ .

$f'' (x) = 2 (\frac{x^{4} + 1 - 4 x^{4}}{{(x^{4} + 1)}^{2}})$

Subtract $4 x^{4}$ from $x^{4}$ .

$f'' (x) = 2 (\frac{- 3 x^{4} + 1}{{(x^{4} + 1)}^{2}})$

Combine $2$ and $\frac{- 3 x^{4} + 1}{{(x^{4} + 1)}^{2}}$ .

$f'' (x) = \frac{2 (- 3 x^{4} + 1)}{{(x^{4} + 1)}^{2}}$

Simplify.

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Apply the distributive property.

$f'' (x) = \frac{2 (- 3 x^{4}) + 2 \cdot 1}{{(x^{4} + 1)}^{2}}$

Simplify each term.

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Multiply $- 3$ by $2$ .

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

Multiply $2$ by $1$ .

$f'' (x) = \frac{- 6 x^{4} + 2}{{(x^{4} + 1)}^{2}}$

Factor $2$ out of $- 6 x^{4} + 2$ .

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Factor $2$ out of $- 6 x^{4}$ .

$f'' (x) = \frac{2 (- 3 x^{4}) + 2}{{(x^{4} + 1)}^{2}}$

Factor $2$ out of $2$ .

$f'' (x) = \frac{2 (- 3 x^{4}) + 2 (1)}{{(x^{4} + 1)}^{2}}$

Factor $2$ out of $2 (- 3 x^{4}) + 2 (1)$ .

$f'' (x) = \frac{2 (- 3 x^{4} + 1)}{{(x^{4} + 1)}^{2}}$

Factor $- 1$ out of $- 3 x^{4}$ .

$f'' (x) = \frac{2 (- (3 x^{4}) + 1)}{{(x^{4} + 1)}^{2}}$

Rewrite $1$ as $- 1 (- 1)$ .

$f'' (x) = \frac{2 (- (3 x^{4}) - 1 \cdot - 1)}{{(x^{4} + 1)}^{2}}$

Factor $- 1$ out of $- (3 x^{4}) - 1 (- 1)$ .

$f'' (x) = \frac{2 (- (3 x^{4} - 1))}{{(x^{4} + 1)}^{2}}$

Rewrite $- (3 x^{4} - 1)$ as $- 1 (3 x^{4} - 1)$ .

$f'' (x) = \frac{2 (- 1 (3 x^{4} - 1))}{{(x^{4} + 1)}^{2}}$

Move the negative in front of the fraction.

$f'' (x) = - \frac{2 (3 x^{4} - 1)}{{(x^{4} + 1)}^{2}}$

The second derivative of $h (x)$ with respect to $x$ is $- \frac{2 (3 x^{4} - 1)}{{(x^{4} + 1)}^{2}}$ .

$- \frac{2 (3 x^{4} - 1)}{{(x^{4} + 1)}^{2}}$