What is the d/(dt)(t-t^{-1}) ?

The d/(dt)(t-t^{-1}) is 1+1/(t^2)

What is the first d/(dt)(t-t^{-1}) ?

The first d/(dt)(t-t^{-1}) is 1+1/(t^2)

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

1 + \frac{1}{t ^{2}}

Solution steps

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

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d t}{d t} - \frac{d}{d t} (t^{- 1})

\frac{d t}{d t} = 1

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

= 1 - (- \frac{1}{t ^{2}})

Simplify

= 1 + \frac{1}{t ^{2}}

2 x^{2} + 7 x^{2} + 4 x

\frac{d}{d x} (7 arcsin (x^{4}))

5 - x^{2}, 4 x^{2}

\frac{d}{d x} (ln (x^{- 3}))

y = \frac{8}{x ^{6}} - \frac{7}{x}