Calculus Examples

$f (x) = \frac{1 - \cos (x)}{\sin (x)}$

Step 1

Find the first derivative.

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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 $f (x) = 1 - \cos (x)$ and $g (x) = \sin (x)$ .

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

Differentiate.

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By the Sum Rule, the derivative of $1 - \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$ .

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

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

$f' (x) = \frac{\sin (x) (0 + \frac{d}{d x} (- \cos (x))) - (1 - \cos (x)) \frac{d}{d x} \sin (x)}{\sin^{2} (x)}$

Add $0$ and $\frac{d}{d x} [- \cos (x)]$ .

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

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

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

Multiply.

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Multiply $- 1$ by $- 1$ .

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

Multiply $\sin (x)$ by $1$ .

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

Raise $\sin (x)$ to the power of $1$ .

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

Raise $\sin (x)$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$f' (x) = \frac{\sin^{2} (x) - (1 - \cos (x)) \cos (x)}{\sin^{2} (x)}$

Simplify.

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Apply the distributive property.

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

Apply the distributive property.

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

Simplify the numerator.

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Simplify each term.

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Multiply $- 1$ by $1$ .

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

Rewrite $- 1 \cos (x)$ as $- \cos (x)$ .

$f' (x) = \frac{\sin^{2} (x) - \cos (x) + \cos (x) \cos (x)}{\sin^{2} (x)}$

Multiply $- - \cos (x)$ .

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Multiply $- 1$ by $- 1$ .

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

Multiply $\cos (x)$ by $1$ .

$f' (x) = \frac{\sin^{2} (x) - \cos (x) + \cos (x) \cos (x)}{\sin^{2} (x)}$

Multiply $\cos (x) \cos (x)$ .

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Raise $\cos (x)$ to the power of $1$ .

$f' (x) = \frac{\sin^{2} (x) - \cos (x) + \cos (x) \cos (x)}{\sin^{2} (x)}$

Raise $\cos (x)$ to the power of $1$ .

$f' (x) = \frac{\sin^{2} (x) - \cos (x) + \cos (x) \cos (x)}{\sin^{2} (x)}$

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

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

Add $1$ and $1$ .

$f' (x) = \frac{\sin^{2} (x) - \cos (x) + \cos^{2} (x)}{\sin^{2} (x)}$

Move $\cos^{2} (x)$ .

$f' (x) = \frac{\sin^{2} (x) + \cos^{2} (x) - \cos (x)}{\sin^{2} (x)}$

Apply pythagorean identity.

$f' (x) = \frac{1 - \cos (x)}{\sin^{2} (x)}$

Step 2

Find the second derivative.

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

Differentiate.

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

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

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

Multiply $2$ by $2$ .

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

By the Sum Rule, the derivative of $1 - \cos (x)$ with respect to $x$ is $\frac{d}{d x} [1] + \frac{d}{d x} [- \cos (x)]$ .

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

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

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

Add $0$ and $\frac{d}{d x} [- \cos (x)]$ .

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

Since $- 1$ is constant with respect to $x$ , the derivative of $- \cos (x)$ with respect to $x$ is $- \frac{d}{d x} [\cos (x)]$ .

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

The derivative of $\cos (x)$ with respect to $x$ is $- \sin (x)$ .

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

Multiply.

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Multiply $- 1$ by $- 1$ .

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

Multiply $\sin (x)$ by $1$ .

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

Multiply $\sin^{2} (x)$ by $\sin (x)$ by adding the exponents.

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Multiply $\sin^{2} (x)$ by $\sin (x)$ .

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Raise $\sin (x)$ to the power of $1$ .

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

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

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

Add $2$ and $1$ .

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

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 $g (x) = \sin (x)$ .

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To apply the Chain Rule, set $u$ as $\sin (x)$ .

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

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

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

Replace all occurrences of $u$ with $\sin (x)$ .

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

Simplify with factoring out.

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Multiply $2$ by $- 1$ .

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

Factor $\sin (x)$ out of $\sin^{3} (x) - 2 (1 - \cos (x)) \sin (x) \frac{d}{d x} [\sin (x)]$ .

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Factor $\sin (x)$ out of $\sin^{3} (x)$ .

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

Factor $\sin (x)$ out of $- 2 (1 - \cos (x)) \sin (x) \frac{d}{d x} [\sin (x)]$ .

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

Factor $\sin (x)$ out of $\sin (x) \sin^{2} (x) + \sin (x) (- 2 (1 - \cos (x)) \frac{d}{d x} [\sin (x)])$ .

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

Cancel the common factors.

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Factor $\sin (x)$ out of $\sin^{4} (x)$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

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

Simplify.

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Apply the distributive property.

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

Apply the distributive property.

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

Simplify each term.

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Multiply $- 2$ by $1$ .

$f'' (x) = \frac{\sin^{2} (x) - 2 \cos (x) - 2 (- \cos (x)) \cos (x)}{\sin^{3} (x)}$

Multiply $- 1$ by $- 2$ .

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

Multiply $2 \cos (x) \cos (x)$ .

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Raise $\cos (x)$ to the power of $1$ .

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

Raise $\cos (x)$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

Step 3

The second derivative of $f (x)$ with respect to $x$ is $\frac{\sin^{2} (x) - 2 \cos (x) + 2 \cos^{2} (x)}{\sin^{3} (x)}$ .

$\frac{\sin^{2} (x) - 2 \cos (x) + 2 \cos^{2} (x)}{\sin^{3} (x)}$