Calculus Examples

$\frac{x}{x^{2} + 16}$

Step 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 $g (x) = x^{2} + 16$ .

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

Multiply $x^{2} + 16$ by $1$ .

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

Step 2.3

By the Sum Rule, the derivative of $x^{2} + 16$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [16]$ .

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

Step 2.4

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

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

Step 2.5

Since $16$ is constant with respect to $x$ , the derivative of $16$ with respect to $x$ is $0$ .

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

Step 2.6

Simplify the expression.

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Step 2.6.1

Add $2 x$ and $0$ .

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

Step 2.6.2

Multiply $2$ by $- 1$ .

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

Step 3

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

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

Step 4

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

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

Step 5

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

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

Step 6

Add $1$ and $1$ .

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

Step 7

Subtract $2 x^{2}$ from $x^{2}$ .

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