Calculus Examples

$\frac{x}{x + 1}$

Step 1

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

$f (x) = \frac{x}{x + 1}$

Step 2

Find the $x$ values where the second derivative is equal to $0$ .

Tap for more steps...

Step 2.1

Find the second derivative.

Tap for more steps...

Step 2.1.1

Find the first derivative.

Tap for more steps...

Step 2.1.1.1

Differentiate using the Quotient Rule which states that $\frac{d}{d x} [\frac{f (x)}{g (x)}]$ is $\frac{g (x) \frac{d}{d x} [f (x)] - f (x) \frac{d}{d x} [g (x)]}{{g (x)}^{2}}$ where $f (x) = x$ and $g (x) = x + 1$ .

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

Step 2.1.1.2

Differentiate.

Tap for more steps...

Step 2.1.1.2.1

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

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

Step 2.1.1.2.2

Multiply $x + 1$ by $1$ .

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

Step 2.1.1.2.3

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

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

Step 2.1.1.2.4

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

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

Step 2.1.1.2.5

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

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

Step 2.1.1.2.6

Simplify by adding terms.

Tap for more steps...

Step 2.1.1.2.6.1

Add $1$ and $0$ .

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

Step 2.1.1.2.6.2

Multiply $- 1$ by $1$ .

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

Step 2.1.1.2.6.3

Subtract $x$ from $x$ .

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

Step 2.1.1.2.6.4

Add $0$ and $1$ .

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

Step 2.1.2

Find the second derivative.

Tap for more steps...

Step 2.1.2.1

Apply basic rules of exponents.

Tap for more steps...

Step 2.1.2.1.1

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

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

Step 2.1.2.1.2

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

Tap for more steps...

Step 2.1.2.1.2.1

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

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

Step 2.1.2.1.2.2

Multiply $2$ by $- 1$ .

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

Step 2.1.2.2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 2}$ and $g (x) = x + 1$ .

Tap for more steps...

Step 2.1.2.2.1

To apply the Chain Rule, set $u$ as $x + 1$ .

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

Step 2.1.2.2.2

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

$- 2 u^{- 3} \frac{d}{d x} [x + 1]$

Step 2.1.2.2.3

Replace all occurrences of $u$ with $x + 1$ .

$- 2 {(x + 1)}^{- 3} \frac{d}{d x} [x + 1]$

Step 2.1.2.3

Differentiate.

Tap for more steps...

Step 2.1.2.3.1

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

$- 2 {(x + 1)}^{- 3} (\frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 2.1.2.3.2

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

$- 2 {(x + 1)}^{- 3} (1 + \frac{d}{d x} [1])$

Step 2.1.2.3.3

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

$- 2 {(x + 1)}^{- 3} (1 + 0)$

Step 2.1.2.3.4

Simplify the expression.

Tap for more steps...

Step 2.1.2.3.4.1

Add $1$ and $0$ .

$- 2 {(x + 1)}^{- 3} \cdot 1$

Step 2.1.2.3.4.2

Multiply $- 2$ by $1$ .

$- 2 {(x + 1)}^{- 3}$

Step 2.1.2.4

Simplify.

Tap for more steps...

Step 2.1.2.4.1

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

$- 2 \frac{1}{{(x + 1)}^{3}}$

Step 2.1.2.4.2

Combine terms.

Tap for more steps...

Step 2.1.2.4.2.1

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

$\frac{- 2}{{(x + 1)}^{3}}$

Step 2.1.2.4.2.2

Move the negative in front of the fraction.

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

Step 2.1.3

The second derivative of $f (x)$ with respect to $x$ is $- \frac{2}{{(x + 1)}^{3}}$ .

$- \frac{2}{{(x + 1)}^{3}}$

Step 2.2

Set the second derivative equal to $0$ then solve the equation $- \frac{2}{{(x + 1)}^{3}} = 0$ .

Tap for more steps...

Step 2.2.1

Set the second derivative equal to $0$ .

$- \frac{2}{{(x + 1)}^{3}} = 0$

Step 2.2.2

Set the numerator equal to zero.

$2 = 0$

Step 2.2.3

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

No solution

Step 3

Find the domain of $f (x) = \frac{x}{x + 1}$ .

Tap for more steps...

Step 3.1

Set the denominator in $\frac{x}{x + 1}$ equal to $0$ to find where the expression is undefined.

$x + 1 = 0$

Step 3.2

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 3.3

The domain is all values of $x$ that make the expression defined.

Interval Notation:

$(- \infty, - 1) \cup (- 1, \infty)$

Set-Builder Notation:

${x | x \neq - 1}$

Interval Notation:

$(- \infty, - 1) \cup (- 1, \infty)$

Set-Builder Notation:

${x | x \neq - 1}$

Step 4

Create intervals around the $x$ -values where the second derivative is zero or undefined.

$(- \infty, - 1) \cup (- 1, \infty)$

Step 5

Substitute any number from the interval $(- \infty, - 1)$ into the second derivative and evaluate to determine the concavity.

Tap for more steps...

Step 5.1

Replace the variable $x$ with $- 4$ in the expression.

$f'' (- 4) = - \frac{2}{{((- 4) + 1)}^{3}}$

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

Add $- 4$ and $1$ .

$f'' (- 4) = - \frac{2}{{(- 3)}^{3}}$

Step 5.2.1.2

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

$f'' (- 4) = - \frac{2}{- 27}$

Step 5.2.2

Move the negative in front of the fraction.

$f'' (- 4) = \frac{2}{27}$

Step 5.2.3

The final answer is $\frac{2}{27}$ .

$\frac{2}{27}$

Step 5.3

The graph is concave up on the interval $(- \infty, - 1)$ because $f'' (- 4)$ is positive.

Concave up on $(- \infty, - 1)$ since $f'' (x)$ is positive

Step 6

Substitute any number from the interval $(- 1, \infty)$ into the second derivative and evaluate to determine the concavity.

Tap for more steps...

Step 6.1

Replace the variable $x$ with $0$ in the expression.

$f'' (0) = - \frac{2}{{((0) + 1)}^{3}}$

Step 6.2

Simplify the result.

Tap for more steps...

Step 6.2.1

Simplify the denominator.

Tap for more steps...

Step 6.2.1.1

Add $0$ and $1$ .

$f'' (0) = - \frac{2}{1^{3}}$

Step 6.2.1.2

One to any power is one.

$f'' (0) = - \frac{2}{1}$

Step 6.2.2

Simplify the expression.

Tap for more steps...

Step 6.2.2.1

Divide $2$ by $1$ .

$f'' (0) = - 1 \cdot 2$

Step 6.2.2.2

Multiply $- 1$ by $2$ .

$f'' (0) = - 2$

Step 6.2.3

The final answer is $- 2$ .

$- 2$

Step 6.3

The graph is concave down on the interval $(- 1, \infty)$ because $f'' (0)$ is negative.

Concave down on $(- 1, \infty)$ since $f'' (x)$ is negative

Step 7

The graph is concave down when the second derivative is negative and concave up when the second derivative is positive.

Concave up on $(- \infty, - 1)$ since $f'' (x)$ is positive

Concave down on $(- 1, \infty)$ since $f'' (x)$ is negative

Step 8