Calculus Examples

$\frac{1}{x^{2} + 3}$

Step 1

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

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d x} [f (g (x))]$ is $f' (g (x)) g' (x)$ where $f (x) = x^{- 1}$ and $g (x) = x^{2} + 3$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x^{2} + 3$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $x^{2} + 3$ .

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

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $x^{2} + 3$ with respect to $x$ is $\frac{d}{d x} [x^{2}] + \frac{d}{d x} [3]$ .

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

Step 3.2

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

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

Step 3.3

Since $3$ is constant with respect to $x$ , the derivative of $3$ with respect to $x$ is $0$ .

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

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $2 x$ and $0$ .

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

Step 3.4.2

Multiply $2$ by $- 1$ .

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

Step 4

Simplify.

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Step 4.1

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

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

Step 4.2

Combine terms.

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Step 4.2.1

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

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

Step 4.2.2

Move the negative in front of the fraction.

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

Step 4.2.3

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

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

Step 4.2.4

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

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