Calculus Examples

$f (x) = \sqrt{x + 1} + \sin (x)$ , $x = 0$

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 $x$ with $0$ in the expression.

$f (0) = \sqrt{(0) + 1} + \sin (0)$

Step 3.2

Simplify $\sqrt{(0) + 1} + \sin (0)$ .

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Step 3.2.1

Remove parentheses.

$\sqrt{(0) + 1} + \sin (0)$

Step 3.2.2

Simplify each term.

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Step 3.2.2.1

Add $0$ and $1$ .

$\sqrt{1} + \sin (0)$

Step 3.2.2.2

Any root of $1$ is $1$ .

$1 + \sin (0)$

Step 3.2.2.3

The exact value of $\sin (0)$ is $0$ .

$1 + 0$

Step 3.2.3

Add $1$ and $0$ .

$1$

Step 4

Find the derivative and evaluate it at $0$ .

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Step 4.1

Find the derivative of $f (x) = \sqrt{x + 1} + \sin (x)$ .

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Step 4.1.1

By the Sum Rule, the derivative of $\sqrt{x + 1} + \sin (x)$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x + 1}] + \frac{d}{d x} [\sin (x)]$ .

$\frac{d}{d x} [\sqrt{x + 1}] + \frac{d}{d x} [\sin (x)]$

Step 4.1.2

Evaluate $\frac{d}{d x} [\sqrt{x + 1}]$ .

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Step 4.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x + 1}$ as ${(x + 1)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [{(x + 1)}^{\frac{1}{2}}] + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{\frac{1}{2}}$ and $g (x) = x + 1$ .

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Step 4.1.2.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x + 1] + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.2.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x + 1] + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.2.3

Replace all occurrences of $u$ with $x + 1$ .

$\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} \frac{d}{d x} [x + 1] + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.3

By the Sum Rule, the derivative of $x + 1$ with respect to $x$ is $\frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

$\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x] + \frac{d}{d x} [1]) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.4

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

$\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} (1 + \frac{d}{d x} [1]) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.5

Since $1$ is constant with respect to $x$ , the derivative of $1$ with respect to $x$ is $0$ .

$\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.6

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

$\frac{1}{2} {(x + 1)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.7

Combine $- 1$ and $\frac{2}{2}$ .

$\frac{1}{2} {(x + 1)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.8

Combine the numerators over the common denominator.

$\frac{1}{2} {(x + 1)}^{\frac{1 - 1 \cdot 2}{2}} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.9

Simplify the numerator.

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Step 4.1.2.9.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(x + 1)}^{\frac{1 - 2}{2}} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.9.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(x + 1)}^{\frac{- 1}{2}} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.10

Move the negative in front of the fraction.

$\frac{1}{2} {(x + 1)}^{- \frac{1}{2}} (1 + 0) + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.11

Add $1$ and $0$ .

$\frac{1}{2} {(x + 1)}^{- \frac{1}{2}} \cdot 1 + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.12

Combine $\frac{1}{2}$ and ${(x + 1)}^{- \frac{1}{2}}$ .

$\frac{{(x + 1)}^{- \frac{1}{2}}}{2} \cdot 1 + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.13

Multiply $\frac{{(x + 1)}^{- \frac{1}{2}}}{2}$ by $1$ .

$\frac{{(x + 1)}^{- \frac{1}{2}}}{2} + \frac{d}{d x} [\sin (x)]$

Step 4.1.2.14

Move ${(x + 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2 {(x + 1)}^{\frac{1}{2}}} + \frac{d}{d x} [\sin (x)]$

Step 4.1.3

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

$\frac{1}{2 {(x + 1)}^{\frac{1}{2}}} + \cos (x)$

Step 4.2

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

$\frac{1}{2 {((0) + 1)}^{\frac{1}{2}}} + \cos (0)$

Step 4.3

Simplify.

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Step 4.3.1

Simplify each term.

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Step 4.3.1.1

Simplify the denominator.

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Step 4.3.1.1.1

Add $0$ and $1$ .

$\frac{1}{2 \cdot 1^{\frac{1}{2}}} + \cos (0)$

Step 4.3.1.1.2

One to any power is one.

$\frac{1}{2 \cdot 1} + \cos (0)$

Step 4.3.1.2

Multiply $2$ by $1$ .

$\frac{1}{2} + \cos (0)$

Step 4.3.1.3

The exact value of $\cos (0)$ is $1$ .

$\frac{1}{2} + 1$

Step 4.3.2

Simplify the expression.

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Step 4.3.2.1

Write $1$ as a fraction with a common denominator.

$\frac{1}{2} + \frac{2}{2}$

Step 4.3.2.2

Combine the numerators over the common denominator.

$\frac{1 + 2}{2}$

Step 4.3.2.3

Add $1$ and $2$ .

$\frac{3}{2}$

Step 5

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

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

Step 6

Simplify each term.

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Step 6.1

Subtract $0$ from $x$ .

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

Step 6.2

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

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

Step 7