Calculus Examples

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

Step 1

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

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

Step 2

Find the first derivative of the function.

Tap for more steps...

Step 2.1

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 $g (x) = 1 + x$ .

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

Step 2.2

Differentiate.

Tap for more steps...

Step 2.2.1

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

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

Step 2.2.2

Multiply $1 + x$ by $1$ .

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

Step 2.2.3

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

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

Step 2.2.4

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

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

Step 2.2.5

Add $0$ and $\frac{d}{d x} [x]$ .

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

Step 2.2.6

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

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

Step 2.2.7

Simplify by adding terms.

Tap for more steps...

Step 2.2.7.1

Multiply $- 1$ by $1$ .

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

Step 2.2.7.2

Subtract $x$ from $x$ .

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

Step 2.2.7.3

Simplify the expression.

Tap for more steps...

Step 2.2.7.3.1

Add $1$ and $0$ .

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

Step 2.2.7.3.2

Reorder terms.

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

Step 3

Find the second derivative of the function.

Tap for more steps...

Step 3.1

Apply basic rules of exponents.

Tap for more steps...

Step 3.1.1

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

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

Step 3.1.2

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

Tap for more steps...

Step 3.1.2.1

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

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

Step 3.1.2.2

Multiply $2$ by $- 1$ .

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

Step 3.2

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 3.2.1

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

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

Step 3.2.2

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

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

Step 3.2.3

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

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

Step 3.3

Differentiate.

Tap for more steps...

Step 3.3.1

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

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

Step 3.3.2

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

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

Step 3.3.3

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

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

Step 3.3.4

Simplify the expression.

Tap for more steps...

Step 3.3.4.1

Add $1$ and $0$ .

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

Step 3.3.4.2

Multiply $- 2$ by $1$ .

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

Step 3.4

Simplify.

Tap for more steps...

Step 3.4.1

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

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

Step 3.4.2

Combine terms.

Tap for more steps...

Step 3.4.2.1

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

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

Step 3.4.2.2

Move the negative in front of the fraction.

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

Step 4

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

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

Step 5

Since there is no value of $x$ that makes the first derivative equal to $0$ , there are no local extrema.

No Local Extrema

Step 6

No Local Extrema

Step 7