Calculus Examples

$f (θ) = \sqrt{\tan (θ) + \cot (θ)}$

Step 1

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

$\frac{d}{d θ} [{(\tan (θ) + \cot (θ))}^{\frac{1}{2}}]$

Step 2

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

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Step 2.1

To apply the Chain Rule, set $u$ as $\tan (θ) + \cot (θ)$ .

$\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 2.2

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

$\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 2.3

Replace all occurrences of $u$ with $\tan (θ) + \cot (θ)$ .

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{\frac{1}{2} - 1} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 3

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

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 4

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

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 5

Combine the numerators over the common denominator.

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 6

Simplify the numerator.

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Step 6.1

Multiply $- 1$ by $2$ .

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{\frac{1 - 2}{2}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 6.2

Subtract $2$ from $1$ .

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{\frac{- 1}{2}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 7

Differentiate using the Sum Rule.

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Step 7.1

Move the negative in front of the fraction.

$\frac{1}{2} {(\tan (θ) + \cot (θ))}^{- \frac{1}{2}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 7.2

Combine fractions.

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Step 7.2.1

Combine $\frac{1}{2}$ and ${(\tan (θ) + \cot (θ))}^{- \frac{1}{2}}$ .

$\frac{{(\tan (θ) + \cot (θ))}^{- \frac{1}{2}}}{2} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 7.2.2

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

$\frac{1}{2 {(\tan (θ) + \cot (θ))}^{\frac{1}{2}}} \frac{d}{d θ} [\tan (θ) + \cot (θ)]$

Step 7.3

By the Sum Rule, the derivative of $\tan (θ) + \cot (θ)$ with respect to $θ$ is $\frac{d}{d θ} [\tan (θ)] + \frac{d}{d θ} [\cot (θ)]$ .

$\frac{1}{2 {(\tan (θ) + \cot (θ))}^{\frac{1}{2}}} (\frac{d}{d θ} [\tan (θ)] + \frac{d}{d θ} [\cot (θ)])$

Step 8

The derivative of $\tan (θ)$ with respect to $θ$ is $\sec^{2} (θ)$ .

$\frac{1}{2 {(\tan (θ) + \cot (θ))}^{\frac{1}{2}}} (\sec^{2} (θ) + \frac{d}{d θ} [\cot (θ)])$

Step 9

The derivative of $\cot (θ)$ with respect to $θ$ is $- \csc^{2} (θ)$ .

$\frac{1}{2 {(\tan (θ) + \cot (θ))}^{\frac{1}{2}}} (\sec^{2} (θ) - \csc^{2} (θ))$

Step 10

Reorder the factors of $\frac{1}{2 {(\tan (θ) + \cot (θ))}^{\frac{1}{2}}} (\sec^{2} (θ) - \csc^{2} (θ))$ .

$(\sec^{2} (θ) - \csc^{2} (θ)) \frac{1}{2 {(\tan (θ) + \cot (θ))}^{\frac{1}{2}}}$