What is the first d/(dt)(t-(t^3)/3) ?

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

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d/(dt)(t-(t^3)/3)

\frac{d}{d t} (t - \frac{t ^{3}}{3})

1 - t^{2}

Solution steps

\frac{d}{d t} (t - \frac{t ^{3}}{3})

Apply the Sum/Difference Rule:

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

= \frac{d t}{d t} - \frac{d}{d t} (\frac{t ^{3}}{3})

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

\frac{d}{d t} (\frac{t ^{3}}{3}) = t^{2}

= 1 - t^{2}

\int (sec^{2} (x)) (2 tan (x) + 1) d x

\frac{\partial}{\partial x} (\frac{1}{4} x^{2} \cdot y)

\frac{d y}{d x} - \frac{4 y}{x} = \frac{y ^{4}}{x ^{2}}

\frac{\partial}{\partial x} (x e^{(\frac{x}{y})})

f (x) = (\frac{x}{2})^{4}