Calculus Examples

$y = \cos (\sqrt{6 t + 12})$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{6 t + 12}$ as ${(6 t + 12)}^{\frac{1}{2}}$ .

$y = \cos ({(6 t + 12)}^{\frac{1}{2}})$

Step 2

Differentiate both sides of the equation.

$\frac{d}{d t} (y) = \frac{d}{d t} (\cos ({(6 t + 12)}^{\frac{1}{2}}))$

Step 3

The derivative of $y$ with respect to $t$ 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 $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = \cos (t)$ and $g (t) = {(6 t + 12)}^{\frac{1}{2}}$ .

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Step 4.1.1

To apply the Chain Rule, set $u_{1}$ as ${(6 t + 12)}^{\frac{1}{2}}$ .

$\frac{d}{d u_{1}} [\cos (u_{1})] \frac{d}{d t} [{(6 t + 12)}^{\frac{1}{2}}]$

Step 4.1.2

The derivative of $\cos (u_{1})$ with respect to $u_{1}$ is $- \sin (u_{1})$ .

$- \sin (u_{1}) \frac{d}{d t} [{(6 t + 12)}^{\frac{1}{2}}]$

Step 4.1.3

Replace all occurrences of $u_{1}$ with ${(6 t + 12)}^{\frac{1}{2}}$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) \frac{d}{d t} [{(6 t + 12)}^{\frac{1}{2}}]$

Step 4.2

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 $g (t) = 6 t + 12$ .

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Step 4.2.1

To apply the Chain Rule, set $u_{2}$ as $6 t + 12$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{d}{d u_{2}} [{u_{2}}^{\frac{1}{2}}] \frac{d}{d t} [6 t + 12])$

Step 4.2.2

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

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {u_{2}}^{\frac{1}{2} - 1} \frac{d}{d t} [6 t + 12])$

Step 4.2.3

Replace all occurrences of $u_{2}$ with $6 t + 12$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{\frac{1}{2} - 1} \frac{d}{d t} [6 t + 12])$

Step 4.3

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

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} [6 t + 12])$

Step 4.4

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

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} [6 t + 12])$

Step 4.5

Combine the numerators over the common denominator.

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} [6 t + 12])$

Step 4.6

Simplify the numerator.

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Step 4.6.1

Multiply $- 1$ by $2$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{\frac{1 - 2}{2}} \frac{d}{d t} [6 t + 12])$

Step 4.6.2

Subtract $2$ from $1$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{\frac{- 1}{2}} \frac{d}{d t} [6 t + 12])$

Step 4.7

Combine fractions.

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Step 4.7.1

Move the negative in front of the fraction.

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2} {(6 t + 12)}^{- \frac{1}{2}} \frac{d}{d t} [6 t + 12])$

Step 4.7.2

Combine $\frac{1}{2}$ and ${(6 t + 12)}^{- \frac{1}{2}}$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{{(6 t + 12)}^{- \frac{1}{2}}}{2} \frac{d}{d t} [6 t + 12])$

Step 4.7.3

Move ${(6 t + 12)}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$- \sin ({(6 t + 12)}^{\frac{1}{2}}) (\frac{1}{2 {(6 t + 12)}^{\frac{1}{2}}} \frac{d}{d t} [6 t + 12])$

Step 4.7.4

Combine $\frac{1}{2 {(6 t + 12)}^{\frac{1}{2}}}$ and $\sin ({(6 t + 12)}^{\frac{1}{2}})$ .

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} \frac{d}{d t} [6 t + 12]$

Step 4.8

By the Sum Rule, the derivative of $6 t + 12$ with respect to $t$ is $\frac{d}{d t} [6 t] + \frac{d}{d t} [12]$ .

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} (\frac{d}{d t} [6 t] + \frac{d}{d t} [12])$

Step 4.9

Since $6$ is constant with respect to $t$ , the derivative of $6 t$ with respect to $t$ is $6 \frac{d}{d t} [t]$ .

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} (6 \frac{d}{d t} [t] + \frac{d}{d t} [12])$

Step 4.10

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

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} (6 \cdot 1 + \frac{d}{d t} [12])$

Step 4.11

Multiply $6$ by $1$ .

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} (6 + \frac{d}{d t} [12])$

Step 4.12

Since $12$ is constant with respect to $t$ , the derivative of $12$ with respect to $t$ is $0$ .

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} (6 + 0)$

Step 4.13

Simplify terms.

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Step 4.13.1

Add $6$ and $0$ .

$- \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}} \cdot 6$

Step 4.13.2

Multiply $6$ by $- 1$ .

$- 6 \frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}}$

Step 4.13.3

Combine $- 6$ and $\frac{\sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}}$ .

$\frac{- 6 \sin ({(6 t + 12)}^{\frac{1}{2}})}{2 {(6 t + 12)}^{\frac{1}{2}}}$

Step 4.13.4

Factor $2$ out of $- 6 \sin ({(6 t + 12)}^{\frac{1}{2}})$ .

$\frac{2 (- 3 \sin ({(6 t + 12)}^{\frac{1}{2}}))}{2 {(6 t + 12)}^{\frac{1}{2}}}$

Step 4.14

Cancel the common factors.

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Step 4.14.1

Factor $2$ out of $2 {(6 t + 12)}^{\frac{1}{2}}$ .

$\frac{2 (- 3 \sin ({(6 t + 12)}^{\frac{1}{2}}))}{2 ({(6 t + 12)}^{\frac{1}{2}})}$

Step 4.14.2

Cancel the common factor.

$\frac{2 (- 3 \sin ({(6 t + 12)}^{\frac{1}{2}}))}{2 {(6 t + 12)}^{\frac{1}{2}}}$

Step 4.14.3

Rewrite the expression.

$\frac{- 3 \sin ({(6 t + 12)}^{\frac{1}{2}})}{{(6 t + 12)}^{\frac{1}{2}}}$

Step 4.15

Move the negative in front of the fraction.

$- \frac{3 \sin ({(6 t + 12)}^{\frac{1}{2}})}{{(6 t + 12)}^{\frac{1}{2}}}$

Step 5

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

$y' = - \frac{3 \sin ({(6 t + 12)}^{\frac{1}{2}})}{{(6 t + 12)}^{\frac{1}{2}}}$

Step 6

Replace $y'$ with $\frac{d y}{d t}$ .

$\frac{d y}{d t} = - \frac{3 \sin ({(6 t + 12)}^{\frac{1}{2}})}{{(6 t + 12)}^{\frac{1}{2}}}$