Calculus Examples

$g (x) = \frac{x^{2} + 4}{4 - x^{2}}$

Step 1

Find the first derivative.

Tap for more steps...

Step 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 $f (x) = x^{2} + 4$ and $g (x) = 4 - x^{2}$ .

$\frac{(4 - x^{2}) \frac{d}{d x} [x^{2} + 4] - (x^{2} + 4) \frac{d}{d x} [4 - x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2

Differentiate.

Tap for more steps...

Step 1.2.1

By the Sum Rule, the derivative of $x^{2} + 4$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]$ .

$\frac{(4 - x^{2}) (\frac{d}{d x} [x^{2}] + \frac{d}{d x} [4]) - (x^{2} + 4) \frac{d}{d x} [4 - x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.2

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

$\frac{(4 - x^{2}) (2 x + \frac{d}{d x} [4]) - (x^{2} + 4) \frac{d}{d x} [4 - x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.3

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{(4 - x^{2}) (2 x + 0) - (x^{2} + 4) \frac{d}{d x} [4 - x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.4

Simplify the expression.

Tap for more steps...

Step 1.2.4.1

Add $2 x$ and $0$ .

$\frac{(4 - x^{2}) (2 x) - (x^{2} + 4) \frac{d}{d x} [4 - x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.4.2

Move $2$ to the left of $4 - x^{2}$ .

$\frac{2 \cdot (4 - x^{2}) x - (x^{2} + 4) \frac{d}{d x} [4 - x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.5

By the Sum Rule, the derivative of $4 - x^{2}$ with respect to $x$ is $\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}]$ .

$\frac{2 (4 - x^{2}) x - (x^{2} + 4) (\frac{d}{d x} [4] + \frac{d}{d x} [- x^{2}])}{{(4 - x^{2})}^{2}}$

Step 1.2.6

Since $4$ is constant with respect to $x$ , the derivative of $4$ with respect to $x$ is $0$ .

$\frac{2 (4 - x^{2}) x - (x^{2} + 4) (0 + \frac{d}{d x} [- x^{2}])}{{(4 - x^{2})}^{2}}$

Step 1.2.7

Add $0$ and $\frac{d}{d x} [- x^{2}]$ .

$\frac{2 (4 - x^{2}) x - (x^{2} + 4) \frac{d}{d x} [- x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.8

Since $- 1$ is constant with respect to $x$ , the derivative of $- x^{2}$ with respect to $x$ is $- \frac{d}{d x} [x^{2}]$ .

$\frac{2 (4 - x^{2}) x - (x^{2} + 4) (- \frac{d}{d x} [x^{2}])}{{(4 - x^{2})}^{2}}$

Step 1.2.9

Multiply.

Tap for more steps...

Step 1.2.9.1

Multiply $- 1$ by $- 1$ .

$\frac{2 (4 - x^{2}) x + 1 (x^{2} + 4) \frac{d}{d x} [x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.9.2

Multiply $x^{2} + 4$ by $1$ .

$\frac{2 (4 - x^{2}) x + (x^{2} + 4) \frac{d}{d x} [x^{2}]}{{(4 - x^{2})}^{2}}$

Step 1.2.10

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

$\frac{2 (4 - x^{2}) x + (x^{2} + 4) (2 x)}{{(4 - x^{2})}^{2}}$

Step 1.2.11

Move $2$ to the left of $x^{2} + 4$ .

$\frac{2 (4 - x^{2}) x + 2 \cdot (x^{2} + 4) x}{{(4 - x^{2})}^{2}}$

Step 1.3

Simplify.

Tap for more steps...

Step 1.3.1

Apply the distributive property.

$\frac{(2 \cdot 4 + 2 (- x^{2})) x + 2 (x^{2} + 4) x}{{(4 - x^{2})}^{2}}$

Step 1.3.2

Apply the distributive property.

$\frac{2 \cdot 4 x + 2 (- x^{2}) x + 2 (x^{2} + 4) x}{{(4 - x^{2})}^{2}}$

Step 1.3.3

Apply the distributive property.

$\frac{2 \cdot 4 x + 2 (- x^{2}) x + (2 x^{2} + 2 \cdot 4) x}{{(4 - x^{2})}^{2}}$

Step 1.3.4

Apply the distributive property.

$\frac{2 \cdot 4 x + 2 (- x^{2}) x + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5

Simplify the numerator.

Tap for more steps...

Step 1.3.5.1

Simplify each term.

Tap for more steps...

Step 1.3.5.1.1

Multiply $2$ by $4$ .

$\frac{8 x + 2 (- x^{2}) x + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.2

Multiply $x^{2}$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.3.5.1.2.1

Move $x$ .

$\frac{8 x + 2 (- (x \cdot x^{2})) + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.2.2

Multiply $x$ by $x^{2}$ .

Tap for more steps...

Step 1.3.5.1.2.2.1

Raise $x$ to the power of $1$ .

$\frac{8 x + 2 (- (x^{1} x^{2})) + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.2.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{8 x + 2 (- x^{1 + 2}) + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.2.3

Add $1$ and $2$ .

$\frac{8 x + 2 (- x^{3}) + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.3

Multiply $- 1$ by $2$ .

$\frac{8 x - 2 x^{3} + 2 x^{2} x + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.4

Multiply $x^{2}$ by $x$ by adding the exponents.

Tap for more steps...

Step 1.3.5.1.4.1

Move $x$ .

$\frac{8 x - 2 x^{3} + 2 (x \cdot x^{2}) + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.4.2

Multiply $x$ by $x^{2}$ .

Tap for more steps...

Step 1.3.5.1.4.2.1

Raise $x$ to the power of $1$ .

$\frac{8 x - 2 x^{3} + 2 (x^{1} x^{2}) + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.4.2.2

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{8 x - 2 x^{3} + 2 x^{1 + 2} + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.4.3

Add $1$ and $2$ .

$\frac{8 x - 2 x^{3} + 2 x^{3} + 2 \cdot 4 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.1.5

Multiply $2$ by $4$ .

$\frac{8 x - 2 x^{3} + 2 x^{3} + 8 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.2

Combine the opposite terms in $8 x - 2 x^{3} + 2 x^{3} + 8 x$ .

Tap for more steps...

Step 1.3.5.2.1

Add $- 2 x^{3}$ and $2 x^{3}$ .

$\frac{8 x + 0 + 8 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.2.2

Add $8 x$ and $0$ .

$\frac{8 x + 8 x}{{(4 - x^{2})}^{2}}$

Step 1.3.5.3

Add $8 x$ and $8 x$ .

$\frac{16 x}{{(4 - x^{2})}^{2}}$

Step 1.3.6

Reorder terms.

$\frac{16 x}{{(- x^{2} + 4)}^{2}}$

Step 1.3.7

Simplify the denominator.

Tap for more steps...

Step 1.3.7.1

Rewrite $4$ as $2^{2}$ .

$\frac{16 x}{{(- x^{2} + 2^{2})}^{2}}$

Step 1.3.7.2

Reorder $- x^{2}$ and $2^{2}$ .

$\frac{16 x}{{(2^{2} - x^{2})}^{2}}$

Step 1.3.7.3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 2$ and $b = x$ .

$\frac{16 x}{{((2 + x) (2 - x))}^{2}}$

Step 1.3.7.4

Apply the product rule to $(2 + x) (2 - x)$ .

$f' (x) = \frac{16 x}{{(2 + x)}^{2} {(2 - x)}^{2}}$

Step 2

Find the second derivative.

Tap for more steps...

Step 2.1

Since $16$ is constant with respect to $x$ , the derivative of $\frac{16 x}{{(2 + x)}^{2} {(2 - x)}^{2}}$ with respect to $x$ is $16 \frac{d}{d x} [\frac{x}{{(2 + x)}^{2} {(2 - x)}^{2}}]$ .

$16 \frac{d}{d x} [\frac{x}{{(2 + x)}^{2} {(2 - x)}^{2}}]$

Step 2.2

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} \frac{d}{d x} [x] - x \frac{d}{d x} [{(2 + x)}^{2} {(2 - x)}^{2}]}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.3

Differentiate using the Power Rule.

Tap for more steps...

Step 2.3.1

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} \cdot 1 - x \frac{d}{d x} [{(2 + x)}^{2} {(2 - x)}^{2}]}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.3.2

Multiply ${(2 + x)}^{2}$ by $1$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x \frac{d}{d x} [{(2 + x)}^{2} {(2 - x)}^{2}]}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.4

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x ({(2 + x)}^{2} \frac{d}{d x} [{(2 - x)}^{2}] + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.5

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 $g (x) = 2 - x$ .

Tap for more steps...

Step 2.5.1

To apply the Chain Rule, set $u_{1}$ as $2 - x$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x ({(2 + x)}^{2} (\frac{d}{d u_{1}} [{u_{1}}^{2}] \frac{d}{d x} [2 - x]) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.5.2

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x ({(2 + x)}^{2} (2 u_{1} \frac{d}{d x} [2 - x]) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.5.3

Replace all occurrences of $u_{1}$ with $2 - x$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x ({(2 + x)}^{2} (2 (2 - x) \frac{d}{d x} [2 - x]) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6

Differentiate.

Tap for more steps...

Step 2.6.1

Move $2$ to the left of ${(2 + x)}^{2}$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (2 \cdot {(2 + x)}^{2} (2 - x) \frac{d}{d x} [2 - x] + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.2

By the Sum Rule, the derivative of $2 - x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [- x]$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (2 {(2 + x)}^{2} (2 - x) (\frac{d}{d x} [2] + \frac{d}{d x} [- x]) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (2 {(2 + x)}^{2} (2 - x) (0 + \frac{d}{d x} [- x]) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.4

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (2 {(2 + x)}^{2} (2 - x) \frac{d}{d x} [- x] + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.5

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (2 {(2 + x)}^{2} (2 - x) (- \frac{d}{d x} [x]) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.6

Multiply $- 1$ by $2$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) \frac{d}{d x} [x] + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.7

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) \cdot 1 + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.6.8

Multiply $- 2$ by $1$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + {(2 - x)}^{2} \frac{d}{d x} [{(2 + x)}^{2}])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.7

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 $g (x) = 2 + x$ .

Tap for more steps...

Step 2.7.1

To apply the Chain Rule, set $u_{2}$ as $2 + x$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + {(2 - x)}^{2} (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d x} [2 + x]))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.7.2

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + {(2 - x)}^{2} (2 u_{2} \frac{d}{d x} [2 + x]))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.7.3

Replace all occurrences of $u_{2}$ with $2 + x$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + {(2 - x)}^{2} (2 (2 + x) \frac{d}{d x} [2 + x]))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8

Differentiate.

Tap for more steps...

Step 2.8.1

Move $2$ to the left of ${(2 - x)}^{2}$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 \cdot {(2 - x)}^{2} (2 + x) \frac{d}{d x} [2 + x])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8.2

By the Sum Rule, the derivative of $2 + x$ with respect to $x$ is $\frac{d}{d x} [2] + \frac{d}{d x} [x]$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x) (\frac{d}{d x} [2] + \frac{d}{d x} [x]))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8.3

Since $2$ is constant with respect to $x$ , the derivative of $2$ with respect to $x$ is $0$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x) (0 + \frac{d}{d x} [x]))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8.4

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x) \frac{d}{d x} [x])}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8.5

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

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x) \cdot 1)}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8.6

Combine fractions.

Tap for more steps...

Step 2.8.6.1

Multiply $2$ by $1$ .

$16 \frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.8.6.2

Combine $16$ and $\frac{{(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$ .

$\frac{16 ({(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x)))}{{({(2 + x)}^{2} {(2 - x)}^{2})}^{2}}$

Step 2.9

Simplify.

Tap for more steps...

Step 2.9.1

Apply the product rule to ${(2 + x)}^{2} {(2 - x)}^{2}$ .

$\frac{16 ({(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x) + 2 {(2 - x)}^{2} (2 + x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.2

Apply the distributive property.

$\frac{16 ({(2 + x)}^{2} {(2 - x)}^{2} - x (- 2 {(2 + x)}^{2} (2 - x)) - x (2 {(2 - x)}^{2} (2 + x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.3

Apply the distributive property.

$\frac{16 ({(2 + x)}^{2} {(2 - x)}^{2}) + 16 (- x (- 2 {(2 + x)}^{2} (2 - x))) + 16 (- x (2 {(2 - x)}^{2} (2 + x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4

Simplify the numerator.

Tap for more steps...

Step 2.9.4.1

Factor $16 (2 + x) (2 - x)$ out of $16 {(2 + x)}^{2} {(2 - x)}^{2} + 16 (- x (- 2 {(2 + x)}^{2} (2 - x))) + 16 (- x (2 {(2 - x)}^{2} (2 + x)))$ .

Tap for more steps...

Step 2.9.4.1.1

Factor $16 (2 + x) (2 - x)$ out of $16 {(2 + x)}^{2} {(2 - x)}^{2}$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x)) + 16 (- x (- 2 {(2 + x)}^{2} (2 - x))) + 16 (- x (2 {(2 - x)}^{2} (2 + x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.1.2

Factor $16 (2 + x) (2 - x)$ out of $16 (- x (- 2 {(2 + x)}^{2} (2 - x)))$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x)) + 16 (2 + x) (2 - x) (- x (- 2 (2 + x))) + 16 (- x (2 {(2 - x)}^{2} (2 + x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.1.3

Factor $16 (2 + x) (2 - x)$ out of $16 (- x (2 {(2 - x)}^{2} (2 + x)))$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x)) + 16 (2 + x) (2 - x) (- x (- 2 (2 + x))) + 16 (2 + x) (2 - x) (- x (2 (2 - x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.1.4

Factor $16 (2 + x) (2 - x)$ out of $16 (2 + x) (2 - x) ((2 + x) (2 - x)) + 16 (2 + x) (2 - x) (- x (- 2 (2 + x)))$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x) - x (- 2 (2 + x))) + 16 (2 + x) (2 - x) (- x (2 (2 - x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.1.5

Factor $16 (2 + x) (2 - x)$ out of $16 (2 + x) (2 - x) ((2 + x) (2 - x) - x (- 2 (2 + x))) + 16 (2 + x) (2 - x) (- x (2 (2 - x)))$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x) - x (- 2 (2 + x)) - x (2 (2 - x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.2

Combine exponents.

Tap for more steps...

Step 2.9.4.2.1

Multiply $- 2$ by $- 1$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x) + 2 x (2 + x) - x (2 (2 - x)))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.2.2

Multiply $2$ by $- 1$ .

$\frac{16 (2 + x) (2 - x) ((2 + x) (2 - x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3

Simplify each term.

Tap for more steps...

Step 2.9.4.3.1

Expand $(2 + x) (2 - x)$ using the FOIL Method.

Tap for more steps...

Step 2.9.4.3.1.1

Apply the distributive property.

$\frac{16 (2 + x) (2 - x) (2 (2 - x) + x (2 - x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.1.2

Apply the distributive property.

$\frac{16 (2 + x) (2 - x) (2 \cdot 2 + 2 (- x) + x (2 - x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.1.3

Apply the distributive property.

$\frac{16 (2 + x) (2 - x) (2 \cdot 2 + 2 (- x) + x \cdot 2 + x (- x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2

Simplify and combine like terms.

Tap for more steps...

Step 2.9.4.3.2.1

Simplify each term.

Tap for more steps...

Step 2.9.4.3.2.1.1

Multiply $2$ by $2$ .

$\frac{16 (2 + x) (2 - x) (4 + 2 (- x) + x \cdot 2 + x (- x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.1.2

Multiply $- 1$ by $2$ .

$\frac{16 (2 + x) (2 - x) (4 - 2 x + x \cdot 2 + x (- x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.1.3

Move $2$ to the left of $x$ .

$\frac{16 (2 + x) (2 - x) (4 - 2 x + 2 \cdot x + x (- x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.1.4

Rewrite using the commutative property of multiplication.

$\frac{16 (2 + x) (2 - x) (4 - 2 x + 2 x - x \cdot x + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.1.5

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 2.9.4.3.2.1.5.1

Move $x$ .

$\frac{16 (2 + x) (2 - x) (4 - 2 x + 2 x - (x \cdot x) + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.1.5.2

Multiply $x$ by $x$ .

$\frac{16 (2 + x) (2 - x) (4 - 2 x + 2 x - x^{2} + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.2

Add $- 2 x$ and $2 x$ .

$\frac{16 (2 + x) (2 - x) (4 + 0 - x^{2} + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.2.3

Add $4$ and $0$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 2 x (2 + x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.3

Apply the distributive property.

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 2 x \cdot 2 + 2 x \cdot x - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.4

Multiply $2$ by $2$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x \cdot x - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.5

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 2.9.4.3.5.1

Move $x$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 (x \cdot x) - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.5.2

Multiply $x$ by $x$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 2 x (2 - x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.6

Apply the distributive property.

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 2 x \cdot 2 - 2 x (- x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.7

Multiply $2$ by $- 2$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 4 x - 2 x (- x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.8

Rewrite using the commutative property of multiplication.

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 4 x - 2 \cdot - 1 x \cdot x)}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.9

Simplify each term.

Tap for more steps...

Step 2.9.4.3.9.1

Multiply $x$ by $x$ by adding the exponents.

Tap for more steps...

Step 2.9.4.3.9.1.1

Move $x$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 4 x - 2 \cdot - 1 (x \cdot x))}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.9.1.2

Multiply $x$ by $x$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 4 x - 2 \cdot - 1 x^{2})}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.3.9.2

Multiply $- 2$ by $- 1$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 4 x + 2 x^{2} - 4 x + 2 x^{2})}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.4

Combine the opposite terms in $4 - x^{2} + 4 x + 2 x^{2} - 4 x + 2 x^{2}$ .

Tap for more steps...

Step 2.9.4.4.1

Subtract $4 x$ from $4 x$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 2 x^{2} + 0 + 2 x^{2})}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.4.2

Add $4 - x^{2} + 2 x^{2}$ and $0$ .

$\frac{16 (2 + x) (2 - x) (4 - x^{2} + 2 x^{2} + 2 x^{2})}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.5

Add $- x^{2}$ and $2 x^{2}$ .

$\frac{16 (2 + x) (2 - x) (4 + x^{2} + 2 x^{2})}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.4.6

Add $x^{2}$ and $2 x^{2}$ .

$\frac{16 (2 + x) (2 - x) (4 + 3 x^{2})}{{({(2 + x)}^{2})}^{2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.5

Combine terms.

Tap for more steps...

Step 2.9.5.1

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

Tap for more steps...

Step 2.9.5.1.1

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

$\frac{16 (2 + x) (2 - x) (4 + 3 x^{2})}{{(2 + x)}^{2 \cdot 2} {({(2 - x)}^{2})}^{2}}$

Step 2.9.5.1.2

Multiply $2$ by $2$ .

$\frac{16 (2 + x) (2 - x) (4 + 3 x^{2})}{{(2 + x)}^{4} {({(2 - x)}^{2})}^{2}}$

Step 2.9.5.2

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

Tap for more steps...

Step 2.9.5.2.1

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

$\frac{16 (2 + x) (2 - x) (4 + 3 x^{2})}{{(2 + x)}^{4} {(2 - x)}^{2 \cdot 2}}$

Step 2.9.5.2.2

Multiply $2$ by $2$ .

$\frac{16 (2 + x) (2 - x) (4 + 3 x^{2})}{{(2 + x)}^{4} {(2 - x)}^{4}}$

Step 2.9.5.3

Cancel the common factor of $2 + x$ and ${(2 + x)}^{4}$ .

Tap for more steps...

Step 2.9.5.3.1

Factor $2 + x$ out of $16 (2 + x) (2 - x) (4 + 3 x^{2})$ .

$\frac{(2 + x) ((16 (2 - x)) (4 + 3 x^{2}))}{{(2 + x)}^{4} {(2 - x)}^{4}}$

Step 2.9.5.3.2

Cancel the common factors.

Tap for more steps...

Step 2.9.5.3.2.1

Factor $2 + x$ out of ${(2 + x)}^{4} {(2 - x)}^{4}$ .

$\frac{(2 + x) ((16 (2 - x)) (4 + 3 x^{2}))}{(2 + x) ({(2 + x)}^{3} {(2 - x)}^{4})}$

Step 2.9.5.3.2.2

Cancel the common factor.

$\frac{(2 + x) ((16 (2 - x)) (4 + 3 x^{2}))}{(2 + x) ({(2 + x)}^{3} {(2 - x)}^{4})}$

Step 2.9.5.3.2.3

Rewrite the expression.

$\frac{(16 (2 - x)) (4 + 3 x^{2})}{{(2 + x)}^{3} {(2 - x)}^{4}}$

Step 2.9.5.4

Cancel the common factor of $2 - x$ and ${(2 - x)}^{4}$ .

Tap for more steps...

Step 2.9.5.4.1

Factor $2 - x$ out of $16 (2 - x) (4 + 3 x^{2})$ .

$\frac{(2 - x) (16 (4 + 3 x^{2}))}{{(2 + x)}^{3} {(2 - x)}^{4}}$

Step 2.9.5.4.2

Cancel the common factors.

Tap for more steps...

Step 2.9.5.4.2.1

Factor $2 - x$ out of ${(2 + x)}^{3} {(2 - x)}^{4}$ .

$\frac{(2 - x) (16 (4 + 3 x^{2}))}{(2 - x) ({(2 + x)}^{3} {(2 - x)}^{3})}$

Step 2.9.5.4.2.2

Cancel the common factor.

$\frac{(2 - x) (16 (4 + 3 x^{2}))}{(2 - x) ({(2 + x)}^{3} {(2 - x)}^{3})}$

Step 2.9.5.4.2.3

Rewrite the expression.

$f'' (x) = \frac{16 (4 + 3 x^{2})}{{(2 + x)}^{3} {(2 - x)}^{3}}$

Step 3

The second derivative of $g (x)$ with respect to $x$ is $\frac{16 (4 + 3 x^{2})}{{(2 + x)}^{3} {(2 - x)}^{3}}$ .

$\frac{16 (4 + 3 x^{2})}{{(2 + x)}^{3} {(2 - x)}^{3}}$