Calculus Examples

$s = \frac{1}{3} t^{3} - 5 t^{2} + 16 t + 8$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d s} (s) = \frac{d}{d s} (\frac{1}{3} t^{3} - 5 t^{2} + 16 t + 8)$

Step 2

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

$1$

Step 3

Differentiate the right side of the equation.

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Step 3.1

By the Sum Rule, the derivative of $\frac{1}{3} t^{3} - 5 t^{2} + 16 t + 8$ with respect to $s$ is $\frac{d}{d s} [\frac{1}{3} t^{3}] + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$ .

$\frac{d}{d s} [\frac{1}{3} t^{3}] + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2

Evaluate $\frac{d}{d s} [\frac{1}{3} t^{3}]$ .

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Step 3.2.1

Since $\frac{1}{3}$ is constant with respect to $s$ , the derivative of $\frac{1}{3} t^{3}$ with respect to $s$ is $\frac{1}{3} \frac{d}{d s} [t^{3}]$ .

$\frac{1}{3} \frac{d}{d s} [t^{3}] + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.2

Differentiate using the chain rule, which states that $\frac{d}{d s} [f (g (s))]$ is $f' (g (s)) g' (s)$ where $f (s) = s^{3}$ and $g (s) = t$ .

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Step 3.2.2.1

To apply the Chain Rule, set $u_{1}$ as $t$ .

$\frac{1}{3} (\frac{d}{d u_{1}} [{u_{1}}^{3}] \frac{d}{d s} [t]) + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.2.2

Differentiate using the Power Rule which states that $\frac{d}{d u_{1}} [{u_{1}}^{n}]$ is $n {u_{1}}^{n - 1}$ where $n = 3$ .

$\frac{1}{3} (3 {u_{1}}^{2} \frac{d}{d s} [t]) + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.2.3

Replace all occurrences of $u_{1}$ with $t$ .

$\frac{1}{3} (3 t^{2} \frac{d}{d s} [t]) + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.3

Rewrite $\frac{d}{d s} [t]$ as $t'$ .

$\frac{1}{3} (3 t^{2} t') + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.4

Combine $3$ and $\frac{1}{3}$ .

$\frac{3}{3} (t^{2} t') + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.5

Combine $t^{2}$ and $\frac{3}{3}$ .

$\frac{t^{2} \cdot 3}{3} t' + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.6

Combine $\frac{t^{2} \cdot 3}{3}$ and $t'$ .

$\frac{t^{2} \cdot 3 t'}{3} + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.7

Move $3$ to the left of $t^{2}$ .

$\frac{3 \cdot t^{2} t'}{3} + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.8

Cancel the common factor of $3$ .

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Step 3.2.8.1

Cancel the common factor.

$\frac{3 t^{2} t'}{3} + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.2.8.2

Divide $t^{2} t'$ by $1$ .

$t^{2} t' + \frac{d}{d s} [- 5 t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.3

Evaluate $\frac{d}{d s} [- 5 t^{2}]$ .

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Step 3.3.1

Since $- 5$ is constant with respect to $s$ , the derivative of $- 5 t^{2}$ with respect to $s$ is $- 5 \frac{d}{d s} [t^{2}]$ .

$t^{2} t' - 5 \frac{d}{d s} [t^{2}] + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.3.2

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

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Step 3.3.2.1

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

$t^{2} t' - 5 (\frac{d}{d u_{2}} [{u_{2}}^{2}] \frac{d}{d s} [t]) + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.3.2.2

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

$t^{2} t' - 5 (2 u_{2} \frac{d}{d s} [t]) + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.3.2.3

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

$t^{2} t' - 5 (2 t \frac{d}{d s} [t]) + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.3.3

Rewrite $\frac{d}{d s} [t]$ as $t'$ .

$t^{2} t' - 5 (2 t t') + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.3.4

Multiply $2$ by $- 5$ .

$t^{2} t' - 10 t t' + \frac{d}{d s} [16 t] + \frac{d}{d s} [8]$

Step 3.4

Evaluate $\frac{d}{d s} [16 t]$ .

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Step 3.4.1

Since $16$ is constant with respect to $s$ , the derivative of $16 t$ with respect to $s$ is $16 \frac{d}{d s} [t]$ .

$t^{2} t' - 10 t t' + 16 \frac{d}{d s} [t] + \frac{d}{d s} [8]$

Step 3.4.2

Rewrite $\frac{d}{d s} [t]$ as $t'$ .

$t^{2} t' - 10 t t' + 16 t' + \frac{d}{d s} [8]$

Step 3.5

Differentiate using the Constant Rule.

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Step 3.5.1

Since $8$ is constant with respect to $s$ , the derivative of $8$ with respect to $s$ is $0$ .

$t^{2} t' - 10 t t' + 16 t' + 0$

Step 3.5.2

Add $t^{2} t' - 10 t t' + 16 t'$ and $0$ .

$t^{2} t' - 10 t t' + 16 t'$

Step 4

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

$1 = t^{2} t' - 10 t t' + 16 t'$

Step 5

Solve for $t'$ .

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Step 5.1

Rewrite the equation as $t^{2} t' - 10 t t' + 16 t' = 1$ .

$t^{2} t' - 10 t t' + 16 t' = 1$

Step 5.2

Factor $t'$ out of $t^{2} t' - 10 t t' + 16 t'$ .

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Step 5.2.1

Factor $t'$ out of $t^{2} t'$ .

$t' t^{2} - 10 t t' + 16 t' = 1$

Step 5.2.2

Factor $t'$ out of $- 10 t t'$ .

$t' (t^{2}) + t' (- 10 t) + 16 t' = 1$

Step 5.2.3

Factor $t'$ out of $16 t'$ .

$t' (t^{2}) + t' (- 10 t) + t' \cdot 16 = 1$

Step 5.2.4

Factor $t'$ out of $t' (t^{2}) + t' (- 10 t)$ .

$t' (t^{2} - 10 t) + t' \cdot 16 = 1$

Step 5.2.5

Factor $t'$ out of $t' (t^{2} - 10 t) + t' \cdot 16$ .

$t' (t^{2} - 10 t + 16) = 1$

Step 5.3

Factor.

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Step 5.3.1

Factor $t^{2} - 10 t + 16$ using the AC method.

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Step 5.3.1.1

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $16$ and whose sum is $- 10$ .

$- 8, - 2$

Step 5.3.1.2

Write the factored form using these integers.

$t' ((t - 8) (t - 2)) = 1$

Step 5.3.2

Remove unnecessary parentheses.

$t' (t - 8) (t - 2) = 1$

Step 5.4

Divide each term in $t' (t - 8) (t - 2) = 1$ by $(t - 8) (t - 2)$ and simplify.

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Step 5.4.1

Divide each term in $t' (t - 8) (t - 2) = 1$ by $(t - 8) (t - 2)$ .

$\frac{t' (t - 8) (t - 2)}{(t - 8) (t - 2)} = \frac{1}{(t - 8) (t - 2)}$

Step 5.4.2

Simplify the left side.

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Step 5.4.2.1

Cancel the common factor of $t - 8$ .

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Step 5.4.2.1.1

Cancel the common factor.

$\frac{t' (t - 8) (t - 2)}{(t - 8) (t - 2)} = \frac{1}{(t - 8) (t - 2)}$

Step 5.4.2.1.2

Rewrite the expression.

$\frac{t' (t - 2)}{t - 2} = \frac{1}{(t - 8) (t - 2)}$

Step 5.4.2.2

Cancel the common factor of $t - 2$ .

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Step 5.4.2.2.1

Cancel the common factor.

$\frac{t' (t - 2)}{t - 2} = \frac{1}{(t - 8) (t - 2)}$

Step 5.4.2.2.2

Divide $t'$ by $1$ .

$t' = \frac{1}{(t - 8) (t - 2)}$

Step 6

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

$\frac{d t}{d s} = \frac{1}{(t - 8) (t - 2)}$