Calculus Examples

$f (x) = \frac{x^{5} - 5 x}{5}$

Find the first derivative of the function.

Tap for more steps...

Since $\frac{1}{5}$ is constant with respect to $x$ , the derivative of $\frac{x^{5} - 5 x}{5}$ with respect to $x$ is $\frac{1}{5} \frac{d}{d x} [x^{5} - 5 x]$ .

$\frac{1}{5} \frac{d}{d x} [x^{5} - 5 x]$

By the Sum Rule, the derivative of $x^{5} - 5 x$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 5 x]$ .

$\frac{1}{5} (\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 5 x])$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$\frac{1}{5} (5 x^{4} + \frac{d}{d x} [- 5 x])$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$\frac{1}{5} (5 x^{4} - 5 \frac{d}{d x} [x])$

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

$\frac{1}{5} (5 x^{4} - 5 \cdot 1)$

Multiply $- 5$ by $1$ .

$\frac{1}{5} (5 x^{4} - 5)$

Simplify.

Tap for more steps...

Apply the distributive property.

$\frac{1}{5} (5 x^{4}) + \frac{1}{5} \cdot - 5$

Combine terms.

Tap for more steps...

Combine $5$ and $\frac{1}{5}$ .

$\frac{5}{5} x^{4} + \frac{1}{5} \cdot - 5$

Combine $\frac{5}{5}$ and $x^{4}$ .

$\frac{5 x^{4}}{5} + \frac{1}{5} \cdot - 5$

Cancel the common factor of $5$ .

Tap for more steps...

Cancel the common factor.

$\frac{5 x^{4}}{5} + \frac{1}{5} \cdot - 5$

Divide $x^{4}$ by $1$ .

$x^{4} + \frac{1}{5} \cdot - 5$

Combine $\frac{1}{5}$ and $- 5$ .

$x^{4} + \frac{- 5}{5}$

Cancel the common factor of $- 5$ and $5$ .

Tap for more steps...

Factor $5$ out of $- 5$ .

$x^{4} + \frac{5 \cdot - 1}{5}$

Cancel the common factors.

Tap for more steps...

Factor $5$ out of $5$ .

$x^{4} + \frac{5 \cdot - 1}{5 (1)}$

Cancel the common factor.

$x^{4} + \frac{5 \cdot - 1}{5 \cdot 1}$

Rewrite the expression.

$x^{4} + \frac{- 1}{1}$

Divide $- 1$ by $1$ .

$x^{4} - 1$

Find the second derivative of the function.

Tap for more steps...

By the Sum Rule, the derivative of $x^{4} - 1$ with respect to $x$ is $\frac{d}{d x} [x^{4}] + \frac{d}{d x} [- 1]$ .

$f'' (x) = \frac{d}{d x} (x^{4}) + \frac{d}{d x} (- 1)$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 4$ .

$f'' (x) = 4 x^{3} + \frac{d}{d x} (- 1)$

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

$f'' (x) = 4 x^{3} + 0$

Add $4 x^{3}$ and $0$ .

$f'' (x) = 4 x^{3}$

To find the local maximum and minimum values of the function, set the derivative equal to $0$ and solve.

$x^{4} - 1 = 0$

Find the first derivative.

Tap for more steps...

Find the first derivative.

Tap for more steps...

Since $\frac{1}{5}$ is constant with respect to $x$ , the derivative of $\frac{x^{5} - 5 x}{5}$ with respect to $x$ is $\frac{1}{5} \frac{d}{d x} [x^{5} - 5 x]$ .

$f' (x) = \frac{1}{5} \cdot \frac{d}{d x} (x^{5} - 5 x)$

By the Sum Rule, the derivative of $x^{5} - 5 x$ with respect to $x$ is $\frac{d}{d x} [x^{5}] + \frac{d}{d x} [- 5 x]$ .

$f' (x) = \frac{1}{5} \cdot (\frac{d}{d x} (x^{5}) + \frac{d}{d x} (- 5 x))$

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 5$ .

$f' (x) = \frac{1}{5} \cdot (5 x^{4} + \frac{d}{d x} (- 5 x))$

Since $- 5$ is constant with respect to $x$ , the derivative of $- 5 x$ with respect to $x$ is $- 5 \frac{d}{d x} [x]$ .

$f' (x) = \frac{1}{5} \cdot (5 x^{4} - 5 \frac{d}{d x} x)$

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

$f' (x) = \frac{1}{5} \cdot (5 x^{4} - 5 \cdot 1)$

Multiply $- 5$ by $1$ .

$f' (x) = \frac{1}{5} \cdot (5 x^{4} - 5)$

Simplify.

Tap for more steps...

Apply the distributive property.

$f' (x) = \frac{1}{5} \cdot (5 x^{4}) + \frac{1}{5} \cdot - 5$

Combine terms.

Tap for more steps...

Combine $5$ and $\frac{1}{5}$ .

$f' (x) = \frac{5}{5} x^{4} + \frac{1}{5} \cdot - 5$

Combine $\frac{5}{5}$ and $x^{4}$ .

$f' (x) = \frac{5 x^{4}}{5} + \frac{1}{5} \cdot - 5$

Cancel the common factor of $5$ .

Tap for more steps...

Cancel the common factor.

$f' (x) = \frac{5 x^{4}}{5} + \frac{1}{5} \cdot - 5$

Divide $x^{4}$ by $1$ .

$f' (x) = x^{4} + \frac{1}{5} \cdot - 5$

Combine $\frac{1}{5}$ and $- 5$ .

$f' (x) = x^{4} + \frac{- 5}{5}$

Cancel the common factor of $- 5$ and $5$ .

Tap for more steps...

Factor $5$ out of $- 5$ .

$f' (x) = x^{4} + \frac{5 \cdot - 1}{5}$

Cancel the common factors.

Tap for more steps...

Factor $5$ out of $5$ .

$f' (x) = x^{4} + \frac{5 \cdot - 1}{5 (1)}$

Cancel the common factor.

$f' (x) = x^{4} + \frac{5 \cdot - 1}{5 \cdot 1}$

Rewrite the expression.

$f' (x) = x^{4} + \frac{- 1}{1}$

Divide $- 1$ by $1$ .

$f' (x) = x^{4} - 1$

The first derivative of $f (x)$ with respect to $x$ is $x^{4} - 1$ .

$x^{4} - 1$

Set the first derivative equal to $0$ then solve the equation $x^{4} - 1 = 0$ .

Tap for more steps...

Set the first derivative equal to $0$ .

$x^{4} - 1 = 0$

Add $1$ to both sides of the equation.

$x^{4} = 1$

Take the 4th root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt[4]{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

The complete solution is the result of both the positive and negative portions of the solution.

Tap for more steps...

First, use the positive value of the $\pm$ to find the first solution.

$x = 1$

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 1$

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, - 1$

Find the values where the derivative is undefined.

Tap for more steps...

The domain of the expression is all real numbers except where the expression is undefined. In this case, there is no real number that makes the expression undefined.

Critical points to evaluate.

$x = 1, - 1$

Evaluate the second derivative at $x = 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4 {(1)}^{3}$

Evaluate the second derivative.

Tap for more steps...

One to any power is one.

$4 \cdot 1$

Multiply $4$ by $1$ .

$4$

$x = 1$ is a local minimum because the value of the second derivative is positive. This is referred to as the second derivative test.

$x = 1$ is a local minimum

Find the y-value when $x = 1$ .

Tap for more steps...

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

$f (1) = \frac{{(1)}^{5} - 5 \cdot 1}{5}$

Simplify the result.

Tap for more steps...

Simplify the numerator.

Tap for more steps...

One to any power is one.

$f (1) = \frac{1 - 5 \cdot 1}{5}$

Multiply $- 5$ by $1$ .

$f (1) = \frac{1 - 5}{5}$

Subtract $5$ from $1$ .

$f (1) = \frac{- 4}{5}$

Move the negative in front of the fraction.

$f (1) = - \frac{4}{5}$

The final answer is $- \frac{4}{5}$ .

$y = - \frac{4}{5}$

Evaluate the second derivative at $x = - 1$ . If the second derivative is positive, then this is a local minimum. If it is negative, then this is a local maximum.

$4 {(- 1)}^{3}$

Evaluate the second derivative.

Tap for more steps...

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

$4 \cdot - 1$

Multiply $4$ by $- 1$ .

$- 4$

$x = - 1$ is a local maximum because the value of the second derivative is negative. This is referred to as the second derivative test.

$x = - 1$ is a local maximum

Find the y-value when $x = - 1$ .

Tap for more steps...

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

$f (- 1) = \frac{{(- 1)}^{5} - 5 \cdot - 1}{5}$

Simplify the result.

Tap for more steps...

Simplify the numerator.

Tap for more steps...

Raise $- 1$ to the power of $5$ .

$f (- 1) = \frac{- 1 - 5 \cdot - 1}{5}$

Multiply $- 5$ by $- 1$ .

$f (- 1) = \frac{- 1 + 5}{5}$

Add $- 1$ and $5$ .

$f (- 1) = \frac{4}{5}$

The final answer is $\frac{4}{5}$ .

$y = \frac{4}{5}$

These are the local extrema for $f (x) = \frac{x^{5} - 5 x}{5}$ .

$(1, - \frac{4}{5})$ is a local minima

$(- 1, \frac{4}{5})$ is a local maxima