Calculus Examples

$f (x) = \sin^{3} (x)$

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

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

$f' (x) = \frac{d}{d u} (u^{3}) \frac{d}{d x} (\sin (x))$

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

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

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

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

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

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

Find the second derivative.

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Since $3$ is constant with respect to $x$ , the derivative of $3 \sin^{2} (x) \cos (x)$ with respect to $x$ is $3 \frac{d}{d x} [\sin^{2} (x) \cos (x)]$ .

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

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

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

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

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

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

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Move $\sin (x)$ .

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

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

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

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

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

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

Add $1$ and $2$ .

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

Simplify the expression.

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Move $- 1$ to the left of $\sin^{3} (x)$ .

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

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

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

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 $\sin (x)$ .

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

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

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

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

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

Move $2$ to the left of $\cos (x)$ .

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

Simplify.

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Apply the distributive property.

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

Combine terms.

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Multiply $- 1$ by $3$ .

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

Multiply $2$ by $3$ .

$f'' (x) = - 3 \sin^{3} (x) + 6 \cos^{2} (x) \sin (x)$

Reorder terms.

$f'' (x) = 6 \cos^{2} (x) \sin (x) - 3 \sin^{3} (x)$

The second derivative of $f (x)$ with respect to $x$ is $6 \cos^{2} (x) \sin (x) - 3 \sin^{3} (x)$ .

$6 \cos^{2} (x) \sin (x) - 3 \sin^{3} (x)$