What is the d/(dt)(1/(t^2)) ?

The d/(dt)(1/(t^2)) is -2/(t^3)

What is the first d/(dt)(1/(t^2)) ?

The first d/(dt)(1/(t^2)) is -2/(t^3)

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\frac{d}{d t} (\frac{1}{t ^{2}})

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

Solution steps

\frac{d}{d t} (\frac{1}{t ^{2}})

Apply exponent rule:

\frac{1}{a} = a^{- 1}

= \frac{d}{d t} (t^{- 2})

Apply the Power Rule:

\frac{d}{d x} (x^{a}) = a \cdot x^{a - 1}

= - 2 t^{- 2 - 1}

Simplify

- 2 t^{- 2 - 1} : - \frac{2}{t ^{3}}

= - \frac{2}{t ^{3}}

\int \frac{5 x - 1}{( 2 x + 1 ) ( x - 1 )} d x

f (x) = arctan (3 x)

\frac{d y}{d t} = k y (100 - y)

\frac{d}{d x} (lo g_{4} ((x^{2} + 2 x)^{3}))

f (x) = x 8 - x^{2}