What is the first d/(dt)(3t^2+2t+1) ?

The first d/(dt)(3t^2+2t+1) is 6t+2

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d/(dt)(3t^2+2t+1)

\frac{d}{d t} (3 t^{2} + 2 t + 1)

6 t + 2

Solution steps

\frac{d}{d t} (3 t^{2} + 2 t + 1)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (3 t^{2}) + \frac{d}{d t} (2 t) + \frac{d}{d t} (1)

\frac{d}{d t} (3 t^{2}) = 6 t

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

\frac{d}{d t} (1) = 0

= 6 t + 2 + 0

Simplify

= 6 t + 2

\frac{x ^{2}}{16}

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

b (1 - sin (x))

cot (x^{2} sin^{2} (x))

a^{3}