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

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

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d/(dt)(t^2-t)

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

2 t - 1

Solution steps

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

Apply the Sum/Difference Rule:

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

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

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

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

= 2 t - 1

\frac{d ^{2}}{d x ^{2}} (2^{x})

x y^{^{'}} + 3 y = 5 x^{2}

\frac{\partial}{\partial t} (e^{- t} cos (π x))

y = (\frac{6 - 5 x}{x ^{2} - 1})^{2}

\int \frac{1}{x ( x - 7 )} d x