Calculus Examples

$- \frac{1}{t^{2}}$

Step 1

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 $f (t) = - 1$ and $g (t) = \frac{1}{t^{2}}$ .

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

Step 2

Differentiate.

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Step 2.1

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

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

Step 2.2

Multiply the exponents in ${(t^{2})}^{- 1}$ .

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Step 2.2.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 2.2.2

Multiply $2$ by $- 1$ .

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

Step 2.3

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

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

Step 2.4

Multiply $- 2$ by $- 1$ .

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

Step 2.5

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

$2 t^{- 3} + \frac{1}{t^{2}} \cdot 0$

Step 2.6

Simplify the expression.

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Step 2.6.1

Multiply $\frac{1}{t^{2}}$ by $0$ .

$2 t^{- 3} + 0$

Step 2.6.2

Add $2 t^{- 3}$ and $0$ .

$2 t^{- 3}$

Step 3

Simplify.

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Step 3.1

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

$2 \frac{1}{t^{3}}$

Step 3.2

Combine $2$ and $\frac{1}{t^{3}}$ .

$\frac{2}{t^{3}}$