Calculus Examples

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

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 $f (x) = x^{2}$ and $g (x) = \sin (2 x)$ .

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To apply the Chain Rule, set $u_{1}$ as $\sin (2 x)$ .

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

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

$f' (x) = 2 u_{1} \frac{d}{d x} (\sin (2 x))$

Replace all occurrences of $u_{1}$ with $\sin (2 x)$ .

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

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

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

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

The derivative of $\sin (u_{2})$ with respect to $u_{2}$ is $\cos (u_{2})$ .

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

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Differentiate.

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

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

Multiply $2$ by $2$ .

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

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

$f' (x) = 4 \sin (2 x) \cos (2 x) \cdot 1$

Simplify the expression.

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Multiply $4$ by $1$ .

$f' (x) = 4 \sin (2 x) \cos (2 x)$

Reorder the factors of $4 \sin (2 x) \cos (2 x)$ .

$f' (x) = 4 \cos (2 x) \sin (2 x)$

Find the second derivative.

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Since $4$ is constant with respect to $x$ , the derivative of $4 \cos (2 x) \sin (2 x)$ with respect to $x$ is $4 \frac{d}{d x} [\cos (2 x) \sin (2 x)]$ .

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

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 $f (x) = \cos (2 x)$ and $g (x) = \sin (2 x)$ .

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

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

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To apply the Chain Rule, set $u_{1}$ as $2 x$ .

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

The derivative of $\sin (u_{1})$ with respect to $u_{1}$ is $\cos (u_{1})$ .

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

Replace all occurrences of $u_{1}$ with $2 x$ .

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

Raise $\cos (2 x)$ to the power of $1$ .

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

Raise $\cos (2 x)$ to the power of $1$ .

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

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

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

Differentiate.

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Add $1$ and $1$ .

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

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

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

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

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

Simplify the expression.

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Multiply $2$ by $1$ .

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

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

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

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

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

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

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Raise $\sin (2 x)$ to the power of $1$ .

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

Raise $\sin (2 x)$ to the power of $1$ .

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

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

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

Add $1$ and $1$ .

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

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

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

Multiply $2$ by $- 1$ .

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

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

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

Multiply $- 2$ by $1$ .

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

Simplify.

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Apply the distributive property.

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

Combine terms.

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Multiply $2$ by $4$ .

$f'' (x) = 8 \cos^{2} (2 x) + 4 (- 2 \sin^{2} (2 x))$

Multiply $- 2$ by $4$ .

$f'' (x) = 8 \cos^{2} (2 x) - 8 \sin^{2} (2 x)$

The second derivative of $f (x)$ with respect to $x$ is $8 \cos^{2} (2 x) - 8 \sin^{2} (2 x)$ .

$8 \cos^{2} (2 x) - 8 \sin^{2} (2 x)$