Calculus Examples

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

Step 1

Write $\frac{1}{x^{2}}$ as a function.

$f (x) = \frac{1}{x^{2}}$

Step 2

Find the first derivative.

Tap for more steps...

Find the first derivative.

Tap for more steps...

Apply basic rules of exponents.

Tap for more steps...

Rewrite $\frac{1}{x^{2}}$ as ${(x^{2})}^{- 1}$ .

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

Multiply the exponents in ${(x^{2})}^{- 1}$ .

Tap for more steps...

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$f' (x) = \frac{d}{d x} (x^{2 \cdot - 1})$

Multiply $2$ by $- 1$ .

$f' (x) = \frac{d}{d x} (x^{- 2})$

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

$f' (x) = - 2 x^{- 3}$

Simplify.

Tap for more steps...

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

$f' (x) = - 2 \frac{1}{x^{3}}$

Combine terms.

Tap for more steps...

Combine $- 2$ and $\frac{1}{x^{3}}$ .

$f' (x) = \frac{- 2}{x^{3}}$

Move the negative in front of the fraction.

$f' (x) = - \frac{2}{x^{3}}$

The first derivative of $f (x)$ with respect to $x$ is $- \frac{2}{x^{3}}$ .

$- \frac{2}{x^{3}}$

Step 3

Set the first derivative equal to $0$ then solve the equation $- \frac{2}{x^{3}} = 0$ .

Tap for more steps...

Set the first derivative equal to $0$ .

$- \frac{2}{x^{3}} = 0$

Set the numerator equal to zero.

$2 = 0$

Since $2 \neq 0$ , there are no solutions.

No solution

Step 4

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 5

Find where the derivative is undefined.

Tap for more steps...

Set the denominator in $\frac{2}{x^{3}}$ equal to $0$ to find where the expression is undefined.

$x^{3} = 0$

Solve for $x$ .

Tap for more steps...

Take the cube root of both sides of the equation to eliminate the exponent on the left side.

$x = \sqrt[3]{0}$

Simplify $\sqrt[3]{0}$ .

Tap for more steps...

Rewrite $0$ as $0^{3}$ .

$x = \sqrt[3]{0^{3}}$

Pull terms out from under the radical, assuming real numbers.

$x = 0$

Step 6

After finding the point that makes the derivative $f' (x) = - \frac{2}{x^{3}}$ equal to $0$ or undefined, the interval to check where $f (x) = \frac{1}{x^{2}}$ is increasing and where it is decreasing is $(- \infty, 0) \cup (0, \infty)$ .

$(- \infty, 0) \cup (0, \infty)$

Step 7

Substitute a value from the interval $(- \infty, 0)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

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

$f' (- 1) = - \frac{2}{{(- 1)}^{3}}$

Simplify the result.

Tap for more steps...

Raise $- 1$ to the power of $3$ .

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

Divide $2$ by $- 1$ .

$f' (- 1) = 2$

The final answer is $2$ .

$2$

At $x = - 1$ the derivative is $2$ . Since this is positive, the function is increasing on $(- \infty, 0)$ .

Increasing on $(- \infty, 0)$ since $f' (x) > 0$

Step 8

Substitute a value from the interval $(0, \infty)$ into the derivative to determine if the function is increasing or decreasing.

Tap for more steps...

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

$f' (1) = - \frac{2}{{(1)}^{3}}$

Simplify the result.

Tap for more steps...

One to any power is one.

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

Divide $2$ by $1$ .

$f' (1) = - 1 \cdot 2$

Multiply $- 1$ by $2$ .

$f' (1) = - 2$

The final answer is $- 2$ .

$- 2$

At $x = 1$ the derivative is $- 2$ . Since this is negative, the function is decreasing on $(0, \infty)$ .

Decreasing on $(0, \infty)$ since $f' (x) < 0$

Step 9

List the intervals on which the function is increasing and decreasing.

Increasing on: $(- \infty, 0)$

Decreasing on: $(0, \infty)$

Step 10