Calculus Examples

$f (x) = \sqrt{x^{2} + 9} - x$

Step 1

Find the first derivative.

Tap for more steps...

Step 1.1

Find the first derivative.

Tap for more steps...

Step 1.1.1

By the Sum Rule, the derivative of $\sqrt{x^{2} + 9} - x$ with respect to $x$ is $\frac{d}{d x} [\sqrt{x^{2} + 9}] + \frac{d}{d x} [- x]$ .

$\frac{d}{d x} [\sqrt{x^{2} + 9}] + \frac{d}{d x} [- x]$

Step 1.1.2

Evaluate $\frac{d}{d x} [\sqrt{x^{2} + 9}]$ .

Tap for more steps...

Step 1.1.2.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x^{2} + 9}$ as ${(x^{2} + 9)}^{\frac{1}{2}}$ .

$\frac{d}{d x} [{(x^{2} + 9)}^{\frac{1}{2}}] + \frac{d}{d x} [- x]$

Step 1.1.2.2

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 $g (x) = x^{2} + 9$ .

Tap for more steps...

Step 1.1.2.2.1

To apply the Chain Rule, set $u$ as $x^{2} + 9$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d x} [x^{2} + 9] + \frac{d}{d x} [- x]$

Step 1.1.2.2.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + 9] + \frac{d}{d x} [- x]$

Step 1.1.2.2.3

Replace all occurrences of $u$ with $x^{2} + 9$ .

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1}{2} - 1} \frac{d}{d x} [x^{2} + 9] + \frac{d}{d x} [- x]$

Step 1.1.2.3

By the Sum Rule, the derivative of $x^{2} + 9$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]$ .

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1}{2} - 1} (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [9]) + \frac{d}{d x} [- x]$

Step 1.1.2.4

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

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1}{2} - 1} (2 x + \frac{d}{d x} [9]) + \frac{d}{d x} [- x]$

Step 1.1.2.5

Since $9$ is constant with respect to $x$ , the derivative of $9$ with respect to $x$ is $0$ .

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1}{2} - 1} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.6

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

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.7

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

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.8

Combine the numerators over the common denominator.

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1 - 1 \cdot 2}{2}} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.9

Simplify the numerator.

Tap for more steps...

Step 1.1.2.9.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(x^{2} + 9)}^{\frac{1 - 2}{2}} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.9.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(x^{2} + 9)}^{\frac{- 1}{2}} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.10

Move the negative in front of the fraction.

$\frac{1}{2} {(x^{2} + 9)}^{- \frac{1}{2}} (2 x + 0) + \frac{d}{d x} [- x]$

Step 1.1.2.11

Add $2 x$ and $0$ .

$\frac{1}{2} {(x^{2} + 9)}^{- \frac{1}{2}} (2 x) + \frac{d}{d x} [- x]$

Step 1.1.2.12

Combine $\frac{1}{2}$ and ${(x^{2} + 9)}^{- \frac{1}{2}}$ .

$\frac{{(x^{2} + 9)}^{- \frac{1}{2}}}{2} (2 x) + \frac{d}{d x} [- x]$

Step 1.1.2.13

Combine $2$ and $\frac{{(x^{2} + 9)}^{- \frac{1}{2}}}{2}$ .

$\frac{2 {(x^{2} + 9)}^{- \frac{1}{2}}}{2} x + \frac{d}{d x} [- x]$

Step 1.1.2.14

Combine $\frac{2 {(x^{2} + 9)}^{- \frac{1}{2}}}{2}$ and $x$ .

$\frac{2 {(x^{2} + 9)}^{- \frac{1}{2}} x}{2} + \frac{d}{d x} [- x]$

Step 1.1.2.15

Move ${(x^{2} + 9)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$\frac{2 x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]$

Step 1.1.2.16

Cancel the common factor.

$\frac{2 x}{2 {(x^{2} + 9)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]$

Step 1.1.2.17

Rewrite the expression.

$\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} + \frac{d}{d x} [- x]$

Step 1.1.3

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

Tap for more steps...

Step 1.1.3.1

Since $- 1$ is constant with respect to $x$ , the derivative of $- x$ with respect to $x$ is $- \frac{d}{d x} [x]$ .

$\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - \frac{d}{d x} [x]$

Step 1.1.3.2

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

$\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1 \cdot 1$

Step 1.1.3.3

Multiply $- 1$ by $1$ .

$f' (x) = \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1$

Step 1.2

The first derivative of $f (x)$ with respect to $x$ is $\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1$ .

$\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1$

Step 2

Set the first derivative equal to $0$ then solve the equation $\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1 = 0$ .

Tap for more steps...

Step 2.1

Set the first derivative equal to $0$ .

$\frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1 = 0$

Step 2.2

Graph each side of the equation. The solution is the x-value of the point of intersection.

No solution

Step 3

There are no values of $x$ in the domain of the original problem where the derivative is $0$ or undefined.

No critical points found

Step 4

No points make the derivative $f' (x) = \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1$ equal to $0$ or undefined. The interval to check if $f (x) = \sqrt{x^{2} + 9} - x$ is increasing or decreasing is $(- \infty, \infty)$ .

$(- \infty, \infty)$

Step 5

Substitute any number, such as $1$ , from the interval $(- \infty, \infty)$ in the derivative $f' (x) = \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1$ to check if the result is negative or positive. If the result is negative, the graph is decreasing on the interval $(- \infty, \infty)$ . If the result is positive, the graph is increasing on the interval $(- \infty, \infty)$ .

Tap for more steps...

Step 5.1

Replace the variable $x$ with $1$ in the expression.

$f' (1) = \frac{1}{{({(1)}^{2} + 9)}^{\frac{1}{2}}} - 1$

Step 5.2

Simplify the result.

Tap for more steps...

Step 5.2.1

Simplify the denominator.

Tap for more steps...

Step 5.2.1.1

One to any power is one.

$f' (1) = \frac{1}{{(1 + 9)}^{\frac{1}{2}}} - 1$

Step 5.2.1.2

Add $1$ and $9$ .

$f' (1) = \frac{1}{10^{\frac{1}{2}}} - 1$

Step 5.2.2

The final answer is $\frac{1}{10^{\frac{1}{2}}} - 1$ .

$\frac{1}{10^{\frac{1}{2}}} - 1$

Step 6

The result of substituting $1$ into $f' (x) = \frac{x}{{(x^{2} + 9)}^{\frac{1}{2}}} - 1$ is $\frac{1}{10^{\frac{1}{2}}} - 1$ , which is negative, so the graph is decreasing on the interval $(- \infty, \infty)$ .

Decreasing on $(- \infty, \infty)$

Step 7

Decreasing over the interval $(- \infty, \infty)$ means that the function is always decreasing.

Always Decreasing

Step 8