Calculus Examples

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

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 x^{3} + x + 1}$ as ${(2 x^{3} + x + 1)}^{\frac{1}{2}}$ .

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

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $2 x^{3} + x + 1$ .

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u$ with $2 x^{3} + x + 1$ .

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

Step 3

To write $- 1$ as a fraction with a common denominator, multiply by $\frac{2}{2}$ .

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

Step 4

Combine $- 1$ and $\frac{2}{2}$ .

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $1$ .

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

Step 7

Combine fractions.

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Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

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

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

Step 7.3

Move ${(2 x^{3} + x + 1)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 8

By the Sum Rule, the derivative of $2 x^{3} + x + 1$ with respect to $x$ is $\frac{d}{d x} [2 x^{3}] + \frac{d}{d x} [x] + \frac{d}{d x} [1]$ .

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

Step 9

Since $2$ is constant with respect to $x$ , the derivative of $2 x^{3}$ with respect to $x$ is $2 \frac{d}{d x} [x^{3}]$ .

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

Step 10

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

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

Step 11

Multiply $3$ by $2$ .

$\frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}} (6 x^{2} + \frac{d}{d x} [x] + \frac{d}{d x} [1])$

Step 12

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

$\frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}} (6 x^{2} + 1 + \frac{d}{d x} [1])$

Step 13

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

$\frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}} (6 x^{2} + 1 + 0)$

Step 14

Add $6 x^{2} + 1$ and $0$ .

$\frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}} (6 x^{2} + 1)$

Step 15

Simplify.

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Step 15.1

Reorder the factors of $\frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}} (6 x^{2} + 1)$ .

$(6 x^{2} + 1) \frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}}$

Step 15.2

Multiply $6 x^{2} + 1$ by $\frac{1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}}$ .

$\frac{6 x^{2} + 1}{2 {(2 x^{3} + x + 1)}^{\frac{1}{2}}}$