Precalculus Examples

$f (x) = \sqrt{- 2 x^{2} + 1}$

Step 1

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

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

Step 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 $g (x) = - 2 x^{2} + 1$ .

Tap for more steps...

Step 2.1

To apply the Chain Rule, set $u$ as $- 2 x^{2} + 1$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $- 2 x^{2} + 1$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

Tap for more steps...

Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $1$ .

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

Step 7

Combine fractions.

Tap for more steps...

Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

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

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

Step 7.3

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

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

Step 8

By the Sum Rule, the derivative of $- 2 x^{2} + 1$ with respect to $x$ is $\frac{d}{d x} [- 2 x^{2}] + \frac{d}{d x} [1]$ .

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

Step 9

Since $- 2$ is constant with respect to $x$ , the derivative of $- 2 x^{2}$ with respect to $x$ is $- 2 \frac{d}{d x} [x^{2}]$ .

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

Step 10

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

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

Step 11

Multiply $2$ by $- 2$ .

$\frac{1}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}} (- 4 x + \frac{d}{d x} [1])$

Step 12

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

$\frac{1}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}} (- 4 x + 0)$

Step 13

Simplify terms.

Tap for more steps...

Step 13.1

Add $- 4 x$ and $0$ .

$\frac{1}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}} (- 4 x)$

Step 13.2

Combine $- 4$ and $\frac{1}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}}$ .

$\frac{- 4}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}} x$

Step 13.3

Combine $\frac{- 4}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}}$ and $x$ .

$\frac{- 4 x}{2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}}$

Step 13.4

Factor $2$ out of $- 4 x$ .

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

Step 14

Cancel the common factors.

Tap for more steps...

Step 14.1

Factor $2$ out of $2 {(- 2 x^{2} + 1)}^{\frac{1}{2}}$ .

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

Step 14.2

Cancel the common factor.

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

Step 14.3

Rewrite the expression.

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

Step 15

Move the negative in front of the fraction.

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