Calculus Examples

$g (t) = \sqrt{2 \cos (5 t)}$

Differentiate using the Constant Multiple Rule.

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Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2 \cos (5 t)}$ as ${(2 \cos (5 t))}^{\frac{1}{2}}$ .

$\frac{d}{d t} [{(2 \cos (5 t))}^{\frac{1}{2}}]$

Simplify with factoring out.

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Factor $2$ out of $2 \cos (5 t)$ .

$\frac{d}{d t} [{(2 (\cos (5 t)))}^{\frac{1}{2}}]$

Apply the product rule to $2 (\cos (5 t))$ .

$\frac{d}{d t} [2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}]$

Since $2^{\frac{1}{2}}$ is constant with respect to $t$ , the derivative of $2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}$ with respect to $t$ is $2^{\frac{1}{2}} \frac{d}{d t} [{\cos (5 t)}^{\frac{1}{2}}]$ .

$2^{\frac{1}{2}} \frac{d}{d t} [{\cos (5 t)}^{\frac{1}{2}}]$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{\frac{1}{2}}$ and $g (t) = \cos (5 t)$ .

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To apply the Chain Rule, set $u_{1}$ as $\cos (5 t)$ .

$2^{\frac{1}{2}} (\frac{d}{d u_{1}} [{u_{1}}^{\frac{1}{2}}] \frac{d}{d t} [\cos (5 t)])$

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

$2^{\frac{1}{2}} (\frac{1}{2} {u_{1}}^{\frac{1}{2} - 1} \frac{d}{d t} [\cos (5 t)])$

Replace all occurrences of $u_{1}$ with $\cos (5 t)$ .

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{\frac{1}{2} - 1} \frac{d}{d t} [\cos (5 t)])$

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

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} [\cos (5 t)])$

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

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} [\cos (5 t)])$

Combine the numerators over the common denominator.

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} [\cos (5 t)])$

Simplify the numerator.

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Multiply $- 1$ by $2$ .

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{\frac{1 - 2}{2}} \frac{d}{d t} [\cos (5 t)])$

Subtract $2$ from $1$ .

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{\frac{- 1}{2}} \frac{d}{d t} [\cos (5 t)])$

Move the negative in front of the fraction.

$2^{\frac{1}{2}} (\frac{1}{2} {\cos (5 t)}^{- \frac{1}{2}} \frac{d}{d t} [\cos (5 t)])$

Combine $\frac{1}{2}$ and ${\cos (5 t)}^{- \frac{1}{2}}$ .

$2^{\frac{1}{2}} (\frac{{\cos (5 t)}^{- \frac{1}{2}}}{2} \frac{d}{d t} [\cos (5 t)])$

Move ${\cos (5 t)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$2^{\frac{1}{2}} (\frac{1}{2 {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)])$

Combine $\frac{1}{2 {\cos (5 t)}^{\frac{1}{2}}}$ and $2^{\frac{1}{2}}$ .

$\frac{2^{\frac{1}{2}}}{2 {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

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

$\frac{1}{2 {\cos (5 t)}^{\frac{1}{2}} \cdot 2^{- \frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Multiply $2$ by $2^{- \frac{1}{2}}$ by adding the exponents.

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Move $2^{- \frac{1}{2}}$ .

$\frac{1}{2^{- \frac{1}{2}} \cdot 2 {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Multiply $2^{- \frac{1}{2}}$ by $2$ .

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Raise $2$ to the power of $1$ .

$\frac{1}{2^{- \frac{1}{2}} \cdot 2^{1} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$\frac{1}{2^{- \frac{1}{2} + 1} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Write $1$ as a fraction with a common denominator.

$\frac{1}{2^{- \frac{1}{2} + \frac{2}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Combine the numerators over the common denominator.

$\frac{1}{2^{\frac{- 1 + 2}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Add $- 1$ and $2$ .

$\frac{1}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [\cos (5 t)]$

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = 5 t$ .

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To apply the Chain Rule, set $u_{2}$ as $5 t$ .

$\frac{1}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} (\frac{d}{d u_{2}} [\cos (u_{2})] \frac{d}{d t} [5 t])$

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

$\frac{1}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} (- \sin (u_{2}) \frac{d}{d t} [5 t])$

Replace all occurrences of $u_{2}$ with $5 t$ .

$\frac{1}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} (- \sin (5 t) \frac{d}{d t} [5 t])$

Differentiate.

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Combine $\frac{1}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}}$ and $\sin (5 t)$ .

$- \frac{\sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [5 t]$

Since $5$ is constant with respect to $t$ , the derivative of $5 t$ with respect to $t$ is $5 \frac{d}{d t} [t]$ .

$- \frac{\sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} (5 \frac{d}{d t} [t])$

Combine fractions.

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Multiply $5$ by $- 1$ .

$- 5 \frac{\sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [t]$

Combine $- 5$ and $\frac{\sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}}$ .

$\frac{- 5 \sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [t]$

Move the negative in front of the fraction.

$- \frac{(5) \sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} \frac{d}{d t} [t]$

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

$- \frac{5 \sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}} \cdot 1$

Multiply $- 1$ by $1$ .

$- \frac{5 \sin (5 t)}{2^{\frac{1}{2}} {\cos (5 t)}^{\frac{1}{2}}}$