What is the first d/(dt)(6t-t^3) ?

The first d/(dt)(6t-t^3) is 6-3t^2

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d/(dt)(6t-t^3)

\frac{d}{d t} (6 t - t^{3})

6 - 3 t^{2}

Solution steps

\frac{d}{d t} (6 t - t^{3})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (6 t) - \frac{d}{d t} (t^{3})

\frac{d}{d t} (6 t) = 6

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

= 6 - 3 t^{2}

(1 - x^{2})^{\frac{1}{2}}

x \to p (x) lim (\frac{f ( x ) - f ( p )}{x - p ( x )})

h \to 0 lim (\frac{sin ( \frac{π}{3} + h ) - sin ( \frac{π}{3} )}{h})

\int \frac{1}{49 e ^{- 6 x} + e ^{6 x}} d x

y^{^{''}} + y = cos^{2} (t), y (0) = 0, y^{^{'}} (0) = 0