What is the first d/(dt)(4+3t+2t^2) ?

The first d/(dt)(4+3t+2t^2) is 4t+3

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d/(dt)(4+3t+2t^2)

\frac{d}{d t} (4 + 3 t + 2 t^{2})

4 t + 3

Solution steps

\frac{d}{d t} (4 + 3 t + 2 t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (4) + \frac{d}{d t} (3 t) + \frac{d}{d t} (2 t^{2})

\frac{d}{d t} (4) = 0

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

\frac{d}{d t} (2 t^{2}) = 4 t

= 0 + 3 + 4 t

Simplify

= 4 t + 3

p + 1

- 3 cos^{3} (x)

\frac{1}{x ^{2} + a ^{2}}

3 e^{2} x

(4 x^{2} - 7)^{\frac{1}{2}}