What is the first d/(dt)(5-6t+t^2) ?

The first d/(dt)(5-6t+t^2) is 2t-6

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d/(dt)(5-6t+t^2)

\frac{d}{d t} (5 - 6 t + t^{2})

2 t - 6

Solution steps

\frac{d}{d t} (5 - 6 t + t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (5) - \frac{d}{d t} (6 t) + \frac{d}{d t} (t^{2})

\frac{d}{d t} (5) = 0

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

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

= 0 - 6 + 2 t

Simplify

= 2 t - 6

\frac{d}{d t} (3 e^{s i n (0.5 t)} - 4)

\frac{d}{d x} (- (cos^{2} (x) + cos^{2} (y)) 2)

\frac{- 1}{( x + 1 )}

\frac{2}{x ^{- 5}}

e^{0.5} t