Calculus Examples

$f (t) = t \sqrt{16 - t^{2}}$

Step 1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{16 - t^{2}}$ as ${(16 - t^{2})}^{\frac{1}{2}}$ .

$\frac{d}{d t} [t {(16 - t^{2})}^{\frac{1}{2}}]$

Step 2

Differentiate using the Product Rule which states that $\frac{d}{d t} [f (t) g (t)]$ is $f (t) \frac{d}{d t} [g (t)] + g (t) \frac{d}{d t} [f (t)]$ where $f (t) = t$ and $g (t) = {(16 - t^{2})}^{\frac{1}{2}}$ .

$t \frac{d}{d t} [{(16 - t^{2})}^{\frac{1}{2}}] + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 3

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 $g (t) = 16 - t^{2}$ .

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Step 3.1

To apply the Chain Rule, set $u$ as $16 - t^{2}$ .

$t (\frac{d}{d u} [u^{\frac{1}{2}}] \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 3.2

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

$t (\frac{1}{2} u^{\frac{1}{2} - 1} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 3.3

Replace all occurrences of $u$ with $16 - t^{2}$ .

$t (\frac{1}{2} {(16 - t^{2})}^{\frac{1}{2} - 1} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 4

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

$t (\frac{1}{2} {(16 - t^{2})}^{\frac{1}{2} - 1 \cdot \frac{2}{2}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 5

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

$t (\frac{1}{2} {(16 - t^{2})}^{\frac{1}{2} + \frac{- 1 \cdot 2}{2}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 6

Combine the numerators over the common denominator.

$t (\frac{1}{2} {(16 - t^{2})}^{\frac{1 - 1 \cdot 2}{2}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 7

Simplify the numerator.

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Step 7.1

Multiply $- 1$ by $2$ .

$t (\frac{1}{2} {(16 - t^{2})}^{\frac{1 - 2}{2}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 7.2

Subtract $2$ from $1$ .

$t (\frac{1}{2} {(16 - t^{2})}^{\frac{- 1}{2}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 8

Combine fractions.

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Step 8.1

Move the negative in front of the fraction.

$t (\frac{1}{2} {(16 - t^{2})}^{- \frac{1}{2}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 8.2

Combine $\frac{1}{2}$ and ${(16 - t^{2})}^{- \frac{1}{2}}$ .

$t (\frac{{(16 - t^{2})}^{- \frac{1}{2}}}{2} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 8.3

Move ${(16 - t^{2})}^{- \frac{1}{2}}$ to the denominator using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$t (\frac{1}{2 {(16 - t^{2})}^{\frac{1}{2}}} \frac{d}{d t} [16 - t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 8.4

Combine $\frac{1}{2 {(16 - t^{2})}^{\frac{1}{2}}}$ and $t$ .

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} \frac{d}{d t} [16 - t^{2}] + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 9

By the Sum Rule, the derivative of $16 - t^{2}$ with respect to $t$ is $\frac{d}{d t} [16] + \frac{d}{d t} [- t^{2}]$ .

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} (\frac{d}{d t} [16] + \frac{d}{d t} [- t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 10

Since $16$ is constant with respect to $t$ , the derivative of $16$ with respect to $t$ is $0$ .

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} (0 + \frac{d}{d t} [- t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 11

Add $0$ and $\frac{d}{d t} [- t^{2}]$ .

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} \frac{d}{d t} [- t^{2}] + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 12

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

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} (- \frac{d}{d t} [t^{2}]) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 13

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

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} (- (2 t)) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 14

Combine fractions.

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Step 14.1

Multiply $2$ by $- 1$ .

$\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}} (- 2 t) + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 14.2

Combine $- 2$ and $\frac{t}{2 {(16 - t^{2})}^{\frac{1}{2}}}$ .

$\frac{- 2 t}{2 {(16 - t^{2})}^{\frac{1}{2}}} t + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 14.3

Combine $\frac{- 2 t}{2 {(16 - t^{2})}^{\frac{1}{2}}}$ and $t$ .

$\frac{- 2 t \cdot t}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 15

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

$\frac{- 2 (t^{1} t)}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 16

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

$\frac{- 2 (t^{1} t^{1})}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 17

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

$\frac{- 2 t^{1 + 1}}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 18

Add $1$ and $1$ .

$\frac{- 2 t^{2}}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 19

Factor $2$ out of $- 2 t^{2}$ .

$\frac{2 (- t^{2})}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 20

Cancel the common factors.

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Step 20.1

Factor $2$ out of $2 {(16 - t^{2})}^{\frac{1}{2}}$ .

$\frac{2 (- t^{2})}{2 ({(16 - t^{2})}^{\frac{1}{2}})} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 20.2

Cancel the common factor.

$\frac{2 (- t^{2})}{2 {(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 20.3

Rewrite the expression.

$\frac{- t^{2}}{{(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 21

Move the negative in front of the fraction.

$- \frac{t^{2}}{{(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \frac{d}{d t} [t]$

Step 22

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

$- \frac{t^{2}}{{(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}} \cdot 1$

Step 23

Multiply ${(16 - t^{2})}^{\frac{1}{2}}$ by $1$ .

$- \frac{t^{2}}{{(16 - t^{2})}^{\frac{1}{2}}} + {(16 - t^{2})}^{\frac{1}{2}}$

Step 24

To write ${(16 - t^{2})}^{\frac{1}{2}}$ as a fraction with a common denominator, multiply by $\frac{{(16 - t^{2})}^{\frac{1}{2}}}{{(16 - t^{2})}^{\frac{1}{2}}}$ .

$- \frac{t^{2}}{{(16 - t^{2})}^{\frac{1}{2}}} + \frac{{(16 - t^{2})}^{\frac{1}{2}} {(16 - t^{2})}^{\frac{1}{2}}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 25

Combine the numerators over the common denominator.

$\frac{- t^{2} + {(16 - t^{2})}^{\frac{1}{2}} {(16 - t^{2})}^{\frac{1}{2}}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 26

Multiply ${(16 - t^{2})}^{\frac{1}{2}}$ by ${(16 - t^{2})}^{\frac{1}{2}}$ by adding the exponents.

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Step 26.1

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

$\frac{- t^{2} + {(16 - t^{2})}^{\frac{1}{2} + \frac{1}{2}}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 26.2

Combine the numerators over the common denominator.

$\frac{- t^{2} + {(16 - t^{2})}^{\frac{1 + 1}{2}}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 26.3

Add $1$ and $1$ .

$\frac{- t^{2} + {(16 - t^{2})}^{\frac{2}{2}}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 26.4

Divide $2$ by $2$ .

$\frac{- t^{2} + {(16 - t^{2})}^{1}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 27

Simplify ${(16 - t^{2})}^{1}$ .

$\frac{- t^{2} + 16 - t^{2}}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 28

Subtract $t^{2}$ from $- t^{2}$ .

$\frac{- 2 t^{2} + 16}{{(16 - t^{2})}^{\frac{1}{2}}}$

Step 29

Reorder terms.

$\frac{- 2 t^{2} + 16}{{(- t^{2} + 16)}^{\frac{1}{2}}}$