Calculus Examples

$f (θ) = \cos (θ + \frac{π}{3})$ , $θ = 0$

Step 1

Consider the function used to find the linearization at $a$ .

$L (x) = f (a) + f' (a) (x - a)$

Step 2

Substitute the value of $a = 0$ into the linearization function.

$L (x) = f (0) + f' (0) (x - 0)$

Step 3

Evaluate $f (0)$ .

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Step 3.1

Replace the variable $θ$ with $0$ in the expression.

$f (0) = \cos ((0) + \frac{π}{3})$

Step 3.2

Simplify $\cos ((0) + \frac{π}{3})$ .

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Step 3.2.1

Remove parentheses.

$\cos ((0) + \frac{π}{3})$

Step 3.2.2

Add $0$ and $\frac{π}{3}$ .

$\cos (\frac{π}{3})$

Step 3.2.3

The exact value of $\cos (\frac{π}{3})$ is $\frac{1}{2}$ .

$\frac{1}{2}$

Step 4

Find the derivative and evaluate it at $0$ .

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Step 4.1

Find the derivative of $f (θ) = \cos (θ + \frac{π}{3})$ .

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Step 4.1.1

Differentiate using the chain rule, which states that $\frac{d}{d θ} [f (g (θ))]$ is $f' (g (θ)) g' (θ)$ where $f (θ) = \cos (θ)$ and $g (θ) = θ + \frac{π}{3}$ .

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Step 4.1.1.1

To apply the Chain Rule, set $u$ as $θ + \frac{π}{3}$ .

$\frac{d}{d u} [\cos (u)] \frac{d}{d θ} [θ + \frac{π}{3}]$

Step 4.1.1.2

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

$- \sin (u) \frac{d}{d θ} [θ + \frac{π}{3}]$

Step 4.1.1.3

Replace all occurrences of $u$ with $θ + \frac{π}{3}$ .

$- \sin (θ + \frac{π}{3}) \frac{d}{d θ} [θ + \frac{π}{3}]$

Step 4.1.2

Differentiate.

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Step 4.1.2.1

By the Sum Rule, the derivative of $θ + \frac{π}{3}$ with respect to $θ$ is $\frac{d}{d θ} [θ] + \frac{d}{d θ} [\frac{π}{3}]$ .

$- \sin (θ + \frac{π}{3}) (\frac{d}{d θ} [θ] + \frac{d}{d θ} [\frac{π}{3}])$

Step 4.1.2.2

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

$- \sin (θ + \frac{π}{3}) (1 + \frac{d}{d θ} [\frac{π}{3}])$

Step 4.1.2.3

Since $\frac{π}{3}$ is constant with respect to $θ$ , the derivative of $\frac{π}{3}$ with respect to $θ$ is $0$ .

$- \sin (θ + \frac{π}{3}) (1 + 0)$

Step 4.1.2.4

Simplify the expression.

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Step 4.1.2.4.1

Add $1$ and $0$ .

$- \sin (θ + \frac{π}{3}) \cdot 1$

Step 4.1.2.4.2

Multiply $- 1$ by $1$ .

$- \sin (θ + \frac{π}{3})$

Step 4.2

Replace the variable $θ$ with $0$ in the expression.

$- \sin ((0) + \frac{π}{3})$

Step 4.3

Simplify.

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Step 4.3.1

Add $0$ and $\frac{π}{3}$ .

$- \sin (\frac{π}{3})$

Step 4.3.2

The exact value of $\sin (\frac{π}{3})$ is $\frac{\sqrt{3}}{2}$ .

$- \frac{\sqrt{3}}{2}$

Step 5

Substitute the components into the linearization function in order to find the linearization at $a$ .

$L (x) = \frac{1}{2} - \frac{\sqrt{3}}{2} (x - 0)$

Step 6

Simplify each term.

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Step 6.1

Subtract $0$ from $x$ .

$L (x) = \frac{1}{2} - \frac{\sqrt{3}}{2} x$

Step 6.2

Combine $x$ and $\frac{\sqrt{3}}{2}$ .

$L (x) = \frac{1}{2} - \frac{x \sqrt{3}}{2}$

Step 7