Calculus Examples

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

Step 1

Find the first derivative.

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Step 1.1

Find the first derivative.

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Step 1.1.1

Apply basic rules of exponents.

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Step 1.1.1.1

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

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

Step 1.1.1.2

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

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Step 1.1.1.2.1

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

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

Step 1.1.1.2.2

Multiply $2$ by $- 1$ .

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

Step 1.1.2

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

$- 2 x^{- 3}$

Step 1.1.3

Simplify.

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Step 1.1.3.1

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

$- 2 \frac{1}{x^{3}}$

Step 1.1.3.2

Combine terms.

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Step 1.1.3.2.1

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

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

Step 1.1.3.2.2

Move the negative in front of the fraction.

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

Step 1.2

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

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

Step 2

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

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Step 2.1

Set the first derivative equal to $0$ .

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

Step 2.2

Set the numerator equal to zero.

$2 = 0$

Step 2.3

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

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

Find where the derivative is undefined.

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Step 4.1

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

$x^{3} = 0$

Step 4.2

Solve for $x$ .

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Step 4.2.1

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

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

Step 4.2.2

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

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Step 4.2.2.1

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

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

Step 4.2.2.2

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

$x = 0$

Step 5

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 6

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

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Step 6.1

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

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

Step 6.2

Simplify the result.

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Step 6.2.1

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

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

Step 6.2.2

Divide $2$ by $- 1$ .

$f' (- 1) = 2$

Step 6.2.3

The final answer is $2$ .

$2$

Step 6.3

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 7

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

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Step 7.1

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

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

Step 7.2

Simplify the result.

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Step 7.2.1

One to any power is one.

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

Step 7.2.2

Divide $2$ by $1$ .

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

Step 7.2.3

Multiply $- 1$ by $2$ .

$f' (1) = - 2$

Step 7.2.4

The final answer is $- 2$ .

$- 2$

Step 7.3

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 8

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

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

Decreasing on: $(0, \infty)$

Step 9