Calculus Examples

f (x) = \sin (x)

$f (x) = \sin (x)$ ,

a = 0

$a = 0$

Step 1

Consider the function used to find the linearization at

a

$a$ .

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

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

Step 2

Substitute the value of

a = 0

$a = 0$ into the linearization function.

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

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

Step 3

Evaluate

f (0)

$f (0)$ .

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Step 3.1

Replace the variable

x

$x$ with

0

$0$ in the expression.

f (0) = \sin (0)

$f (0) = \sin (0)$

Step 3.2

Simplify

\sin (0)

$\sin (0)$ .

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Step 3.2.1

Remove parentheses.

\sin (0)

$\sin (0)$

Step 3.2.2

The exact value of

\sin (0)

$\sin (0)$ is

0

$0$ .

0

$0$

0

$0$

0

$0$

Step 4

Find the derivative and evaluate it at

0

$0$ .

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Step 4.1

The derivative of

\sin (x)

$\sin (x)$ with respect to

x

$x$ is

\cos (x)

$\cos (x)$ .

\cos (x)

$\cos (x)$

Step 4.2

Replace the variable

x

$x$ with

0

$0$ in the expression.

\cos (0)

$\cos (0)$

Step 4.3

The exact value of

\cos (0)

$\cos (0)$ is

1

$1$ .

1

$1$

1

$1$

Step 5

Substitute the components into the linearization function in order to find the linearization at

a

$a$ .

L (x) = 0 + 1 (x - 0)

$L (x) = 0 + 1 (x - 0)$

Step 6

Simplify.

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Step 6.1

Add

0

$0$ and

1 (x - 0)

$1 (x - 0)$ .

L (x) = 1 (x - 0)

$L (x) = 1 (x - 0)$

Step 6.2

Multiply

x - 0

$x - 0$ by

1

$1$ .

L (x) = x - 0

$L (x) = x - 0$

Step 6.3

Subtract

0

$0$ from

x

$x$ .

L (x) = x

$L (x) = x$

L (x) = x

$L (x) = x$

Step 7