Calculus Examples

$s = t^{3} - 12 t$

Step 1

Differentiate both sides of the equation.

$\frac{d}{d s} (s) = \frac{d}{d s} (t^{3} - 12 t)$

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 $t^{3} - 12 t$ with respect to $s$ is $\frac{d}{d s} [t^{3}] + \frac{d}{d s} [- 12 t]$ .

$\frac{d}{d s} [t^{3}] + \frac{d}{d s} [- 12 t]$

Step 3.2

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

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Step 3.2.1

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 3.2.1.1

To apply the Chain Rule, set $u$ as $t$ .

$\frac{d}{d u} [u^{3}] \frac{d}{d s} [t] + \frac{d}{d s} [- 12 t]$

Step 3.2.1.2

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

$3 u^{2} \frac{d}{d s} [t] + \frac{d}{d s} [- 12 t]$

Step 3.2.1.3

Replace all occurrences of $u$ with $t$ .

$3 t^{2} \frac{d}{d s} [t] + \frac{d}{d s} [- 12 t]$

Step 3.2.2

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

$3 t^{2} t' + \frac{d}{d s} [- 12 t]$

Step 3.3

Evaluate $\frac{d}{d s} [- 12 t]$ .

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Step 3.3.1

Since $- 12$ is constant with respect to $s$ , the derivative of $- 12 t$ with respect to $s$ is $- 12 \frac{d}{d s} [t]$ .

$3 t^{2} t' - 12 \frac{d}{d s} [t]$

Step 3.3.2

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

$3 t^{2} t' - 12 t'$

Step 4

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

$1 = 3 t^{2} t' - 12 t'$

Step 5

Solve for $t'$ .

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Step 5.1

Rewrite the equation as $3 t^{2} t' - 12 t' = 1$ .

$3 t^{2} t' - 12 t' = 1$

Step 5.2

Factor $3 t'$ out of $3 t^{2} t' - 12 t'$ .

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Step 5.2.1

Factor $3 t'$ out of $3 t^{2} t'$ .

$3 t' t^{2} - 12 t' = 1$

Step 5.2.2

Factor $3 t'$ out of $- 12 t'$ .

$3 t' t^{2} + 3 t' \cdot - 4 = 1$

Step 5.2.3

Factor $3 t'$ out of $3 t' t^{2} + 3 t' \cdot - 4$ .

$3 t' (t^{2} - 4) = 1$

Step 5.3

Rewrite $4$ as $2^{2}$ .

$3 t' (t^{2} - 2^{2}) = 1$

Step 5.4

Factor.

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Step 5.4.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = t$ and $b = 2$ .

$3 t' ((t + 2) (t - 2)) = 1$

Step 5.4.2

Remove unnecessary parentheses.

$3 t' (t + 2) (t - 2) = 1$

Step 5.5

Divide each term in $3 t' (t + 2) (t - 2) = 1$ by $3 (t + 2) (t - 2)$ and simplify.

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Step 5.5.1

Divide each term in $3 t' (t + 2) (t - 2) = 1$ by $3 (t + 2) (t - 2)$ .

$\frac{3 t' (t + 2) (t - 2)}{3 (t + 2) (t - 2)} = \frac{1}{3 (t + 2) (t - 2)}$

Step 5.5.2

Simplify the left side.

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Step 5.5.2.1

Cancel the common factor of $3$ .

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Step 5.5.2.1.1

Cancel the common factor.

$\frac{3 t' (t + 2) (t - 2)}{3 (t + 2) (t - 2)} = \frac{1}{3 (t + 2) (t - 2)}$

Step 5.5.2.1.2

Rewrite the expression.

$\frac{t' (t + 2) (t - 2)}{(t + 2) (t - 2)} = \frac{1}{3 (t + 2) (t - 2)}$

Step 5.5.2.2

Cancel the common factor of $t + 2$ .

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Step 5.5.2.2.1

Cancel the common factor.

$\frac{t' (t + 2) (t - 2)}{(t + 2) (t - 2)} = \frac{1}{3 (t + 2) (t - 2)}$

Step 5.5.2.2.2

Rewrite the expression.

$\frac{t' (t - 2)}{t - 2} = \frac{1}{3 (t + 2) (t - 2)}$

Step 5.5.2.3

Cancel the common factor of $t - 2$ .

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Step 5.5.2.3.1

Cancel the common factor.

$\frac{t' (t - 2)}{t - 2} = \frac{1}{3 (t + 2) (t - 2)}$

Step 5.5.2.3.2

Divide $t'$ by $1$ .

$t' = \frac{1}{3 (t + 2) (t - 2)}$

Step 6

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

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