Calculus Examples

$y = \sin (2 x) \cos (x^{2})$

Find the first derivative.

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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 $g (x) = \cos (x^{2})$ .

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

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

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

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

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

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

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

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

Differentiate using the Power Rule.

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

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

Multiply $2$ by $- 1$ .

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

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 $2 x$ .

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

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

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

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

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

Differentiate.

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

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

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

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

Simplify the expression.

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Multiply $2$ by $1$ .

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

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

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

Reorder terms.

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

Find the second derivative.

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By the Sum Rule, the derivative of $- 2 x \sin (x^{2}) \sin (2 x) + 2 \cos (x^{2}) \cos (2 x)$ with respect to $x$ is $\frac{d}{d x} [- 2 x \sin (x^{2}) \sin (2 x)] + \frac{d}{d x} [2 \cos (x^{2}) \cos (2 x)]$ .

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

Evaluate $\frac{d}{d x} [- 2 x \sin (x^{2}) \sin (2 x)]$ .

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

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

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

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

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 $2 x$ .

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

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

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

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

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

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

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

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

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

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

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

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 $g (x) = x^{2}$ .

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

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

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

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

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

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

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

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

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

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

Multiply $2$ by $1$ .

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

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

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

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

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

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

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

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

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

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

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

Add $1$ and $1$ .

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

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

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

Multiply $\sin (x^{2})$ by $1$ .

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

Evaluate $\frac{d}{d x} [2 \cos (x^{2}) \cos (2 x)]$ .

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

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

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

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

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 $u_{3}$ as $2 x$ .

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

The derivative of $\cos (u_{3})$ with respect to $u_{3}$ is $- \sin (u_{3})$ .

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

Replace all occurrences of $u_{3}$ with $2 x$ .

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

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

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

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

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

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 $g (x) = x^{2}$ .

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To apply the Chain Rule, set $u_{4}$ as $x^{2}$ .

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

The derivative of $\cos (u_{4})$ with respect to $u_{4}$ is $- \sin (u_{4})$ .

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

Replace all occurrences of $u_{4}$ with $x^{2}$ .

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

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

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

Multiply $2$ by $1$ .

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

Multiply $2$ by $- 1$ .

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

Multiply $2$ by $- 1$ .

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

Simplify.

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Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Combine terms.

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Multiply $2$ by $- 2$ .

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

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

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

Multiply $2$ by $- 2$ .

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

Multiply $- 2$ by $2$ .

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

Multiply $- 2$ by $2$ .

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

Reorder the factors of $- 4 x \sin (x^{2}) \cos (2 x)$ .

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

Subtract $4 \cos (2 x) \sin (x^{2}) x$ from $- 4 x \cos (2 x) \sin (x^{2})$ .

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Move $\cos (2 x)$ .

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

Subtract $4 x \cos (2 x) \sin (x^{2})$ from $- 4 x \cos (2 x) \sin (x^{2})$ .

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

Reorder terms.

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