Precalculus Examples

$f (x) = \sqrt{x} - 2$

Step 1

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

$\frac{d}{d x} [\sqrt{x}] + \frac{d}{d x} [- 2]$

Step 2

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

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

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

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

Step 2.4

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

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

Step 2.5

Combine the numerators over the common denominator.

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

Step 2.6

Simplify the numerator.

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Step 2.6.1

Multiply $- 1$ by $2$ .

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

Step 2.6.2

Subtract $2$ from $1$ .

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

Step 2.7

Move the negative in front of the fraction.

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

Step 3

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2$ with respect to $x$ is $0$ .

$\frac{1}{2} x^{- \frac{1}{2}} + 0$

Step 4

Simplify.

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Step 4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{1}{2} \cdot \frac{1}{x^{\frac{1}{2}}} + 0$

Step 4.2

Combine terms.

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Step 4.2.1

Multiply $\frac{1}{2}$ by $\frac{1}{x^{\frac{1}{2}}}$ .

$\frac{1}{2 x^{\frac{1}{2}}} + 0$

Step 4.2.2

Add $\frac{1}{2 x^{\frac{1}{2}}}$ and $0$ .

$\frac{1}{2 x^{\frac{1}{2}}}$