Calculus Examples

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

Step 1

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

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

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 $g (x) = x^{3} + 1$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $x^{3} + 1$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $x^{3} + 1$ .

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

Step 3

Differentiate.

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Step 3.1

By the Sum Rule, the derivative of $x^{3} + 1$ with respect to $x$ is $\frac{d}{d x} [x^{3}] + \frac{d}{d x} [1]$ .

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

Step 3.2

Differentiate using the Power Rule which states that $\frac{d}{d x} [x^{n}]$ is $n x^{n - 1}$ where $n = 3$ .

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

Step 3.3

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

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

Step 3.4

Simplify the expression.

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Step 3.4.1

Add $3 x^{2}$ and $0$ .

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

Step 3.4.2

Multiply $3$ by $- 1$ .

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

Step 4

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

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

Step 5

Combine terms.

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Step 5.1

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

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

Step 5.2

Move the negative in front of the fraction.

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

Step 5.3

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

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

Step 5.4

Move $3$ to the left of $x^{2}$ .

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