Calculus Examples

$y = \sqrt{\sin^{2} (3 x) + x}$

Step 1

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

$y = {(\sin^{2} (3 x) + x)}^{\frac{1}{2}}$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d x} (y) = \frac{d}{d x} ({(\sin^{2} (3 x) + x)}^{\frac{1}{2}})$

Step 3

The derivative of $y$ with respect to $x$ is $y'$ .

$y'$

Step 4

Differentiate the right side of the equation.

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Step 4.1

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 $g (x) = \sin^{2} (3 x) + x$ .

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Step 4.1.1

To apply the Chain Rule, set $u_{1}$ as $\sin^{2} (3 x) + x$ .

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

Step 4.1.2

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

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

Step 4.1.3

Replace all occurrences of $u_{1}$ with $\sin^{2} (3 x) + x$ .

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

Combine the numerators over the common denominator.

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

Step 4.5

Simplify the numerator.

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Step 4.5.1

Multiply $- 1$ by $2$ .

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

Step 4.5.2

Subtract $2$ from $1$ .

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

Step 4.6

Differentiate using the Sum Rule.

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Step 4.6.1

Move the negative in front of the fraction.

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

Step 4.6.2

Combine fractions.

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Step 4.6.2.1

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

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

Step 4.6.2.2

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

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

Step 4.6.3

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

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

Step 4.7

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 $g (x) = \sin (3 x)$ .

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Step 4.7.1

To apply the Chain Rule, set $u_{2}$ as $\sin (3 x)$ .

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

Step 4.7.2

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

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

Step 4.7.3

Replace all occurrences of $u_{2}$ with $\sin (3 x)$ .

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

Step 4.8

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

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Step 4.8.1

To apply the Chain Rule, set $u_{3}$ as $3 x$ .

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

Step 4.8.2

The derivative of $\sin (u_{3})$ with respect to $u_{3}$ is $\cos (u_{3})$ .

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

Step 4.8.3

Replace all occurrences of $u_{3}$ with $3 x$ .

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

Step 4.9

Differentiate.

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Step 4.9.1

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

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

Step 4.9.2

Multiply $3$ by $2$ .

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

Step 4.9.3

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

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

Step 4.9.4

Multiply $6$ by $1$ .

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

Step 4.9.5

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

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

Step 4.9.6

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

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

Step 5

Reform the equation by setting the left side equal to the right side.

$y' = (6 \sin (3 x) \cos (3 x) + 1) (\frac{1}{2 {(\sin^{2} (3 x) + x)}^{\frac{1}{2}}})$

Step 6

Replace $y'$ with $\frac{d y}{d x}$ .

$\frac{d y}{d x} = (6 \sin (3 x) \cos (3 x) + 1) (\frac{1}{2 {(\sin^{2} (3 x) + x)}^{\frac{1}{2}}})$