Calculus Examples

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

Step 1

Rewrite $\frac{1}{t^{2} + 3 t - 1}$ as ${(t^{2} + 3 t - 1)}^{- 1}$ .

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

Step 2

Differentiate using the chain rule, which states that $\frac{d}{d t} [f (g (t))]$ is $f' (g (t)) g' (t)$ where $f (t) = t^{- 1}$ and $g (t) = t^{2} + 3 t - 1$ .

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Step 2.1

To apply the Chain Rule, set $u$ as $t^{2} + 3 t - 1$ .

$\frac{d}{d u} [u^{- 1}] \frac{d}{d t} [t^{2} + 3 t - 1]$

Step 2.2

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

$- u^{- 2} \frac{d}{d t} [t^{2} + 3 t - 1]$

Step 2.3

Replace all occurrences of $u$ with $t^{2} + 3 t - 1$ .

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

Step 3

Differentiate.

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Step 3.1

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

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

Step 3.2

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

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

Step 3.3

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

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

Step 3.4

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

$- {(t^{2} + 3 t - 1)}^{- 2} (2 t + 3 \cdot 1 + \frac{d}{d t} [- 1])$

Step 3.5

Multiply $3$ by $1$ .

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

Step 3.6

Since $- 1$ is constant with respect to $t$ , the derivative of $- 1$ with respect to $t$ is $0$ .

$- {(t^{2} + 3 t - 1)}^{- 2} (2 t + 3 + 0)$

Step 3.7

Add $2 t + 3$ and $0$ .

$- {(t^{2} + 3 t - 1)}^{- 2} (2 t + 3)$

Step 4

Simplify.

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Step 4.1

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

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

Step 4.2

Reorder the factors of $- \frac{1}{{(t^{2} + 3 t - 1)}^{2}} (2 t + 3)$ .

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

Step 4.3

Apply the distributive property.

$(- (2 t) - 1 \cdot 3) \frac{1}{{(t^{2} + 3 t - 1)}^{2}}$

Step 4.4

Multiply $2$ by $- 1$ .

$(- 2 t - 1 \cdot 3) \frac{1}{{(t^{2} + 3 t - 1)}^{2}}$

Step 4.5

Multiply $- 1$ by $3$ .

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

Step 4.6

Multiply $- 2 t - 3$ by $\frac{1}{{(t^{2} + 3 t - 1)}^{2}}$ .

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

Step 4.7

Factor $- 1$ out of $- 2 t$ .

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

Step 4.8

Rewrite $- 3$ as $- 1 (3)$ .

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

Step 4.9

Factor $- 1$ out of $- (2 t) - 1 (3)$ .

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

Step 4.10

Rewrite $- (2 t + 3)$ as $- 1 (2 t + 3)$ .

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

Step 4.11

Move the negative in front of the fraction.

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