Calculus Examples

$y = \sqrt{\cos (2 x) - \sin (x)}$

Step 1

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

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

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 $g (x) = \cos (2 x) - \sin (x)$ .

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Step 2.1

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

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

Step 2.2

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

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

Step 2.3

Replace all occurrences of $u_{1}$ with $\cos (2 x) - \sin (x)$ .

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

Step 3

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

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

Step 4

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

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

Step 5

Combine the numerators over the common denominator.

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

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

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

Step 6.2

Subtract $2$ from $1$ .

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

Step 7

Differentiate using the Sum Rule.

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Step 7.1

Move the negative in front of the fraction.

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

Step 7.2

Combine fractions.

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Step 7.2.1

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

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

Step 7.2.2

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

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

Step 7.3

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

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

Step 8

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

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Step 8.1

To apply the Chain Rule, set $u_{2}$ as $2 x$ .

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

Step 8.2

The derivative of $\cos (u_{2})$ with respect to $u_{2}$ is $- \sin (u_{2})$ .

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

Step 8.3

Replace all occurrences of $u_{2}$ with $2 x$ .

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

Step 9

Differentiate.

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Step 9.1

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

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

Step 9.2

Multiply $2$ by $- 1$ .

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

Step 9.3

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

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

Step 9.4

Multiply $- 2$ by $1$ .

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

Step 9.5

Since $- 1$ is constant with respect to $x$ , the derivative of $- \sin (x)$ with respect to $x$ is $- \frac{d}{d x} [\sin (x)]$ .

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

Step 10

The derivative of $\sin (x)$ with respect to $x$ is $\cos (x)$ .

$\frac{1}{2 {(\cos (2 x) - \sin (x))}^{\frac{1}{2}}} (- 2 \sin (2 x) - \cos (x))$

Step 11

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

$(- 2 \sin (2 x) - \cos (x)) \frac{1}{2 {(\cos (2 x) - \sin (x))}^{\frac{1}{2}}}$