Calculus Examples

$f (x) = \sin (x)$ , $a = \frac{π}{6}$

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 $a = \frac{π}{6}$ into the linearization function.

$L (x) = f (\frac{π}{6}) + f' (\frac{π}{6}) (x - \frac{π}{6})$

Step 3

Evaluate $f (\frac{π}{6})$ .

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Step 3.1

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$f (\frac{π}{6}) = \sin (\frac{π}{6})$

Step 3.2

Simplify $\sin (\frac{π}{6})$ .

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Step 3.2.1

Remove parentheses.

$\sin (\frac{π}{6})$

Step 3.2.2

The exact value of $\sin (\frac{π}{6})$ is $\frac{1}{2}$ .

$\frac{1}{2}$

Step 4

Find the derivative and evaluate it at $\frac{π}{6}$ .

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Step 4.1

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

$\cos (x)$

Step 4.2

Replace the variable $x$ with $\frac{π}{6}$ in the expression.

$\cos (\frac{π}{6})$

Step 4.3

The exact value of $\cos (\frac{π}{6})$ 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 $a$ .

$L (x) = \frac{1}{2} + \frac{\sqrt{3}}{2} (x - \frac{π}{6})$

Step 6

Simplify each term.

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Step 6.1

Apply the distributive property.

$L (x) = \frac{1}{2} + \frac{\sqrt{3}}{2} x + \frac{\sqrt{3}}{2} \cdot (- \frac{π}{6})$

Step 6.2

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

$L (x) = \frac{1}{2} + \frac{\sqrt{3} x}{2} + \frac{\sqrt{3}}{2} \cdot (- \frac{π}{6})$

Step 6.3

Multiply $\frac{\sqrt{3}}{2} (- \frac{π}{6})$ .

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Step 6.3.1

Multiply $\frac{\sqrt{3}}{2}$ by $\frac{π}{6}$ .

$L (x) = \frac{1}{2} + \frac{\sqrt{3} x}{2} - \frac{\sqrt{3} π}{2 \cdot 6}$

Step 6.3.2

Multiply $2$ by $6$ .

$L (x) = \frac{1}{2} + \frac{\sqrt{3} x}{2} - \frac{\sqrt{3} π}{12}$

Step 7