What is the d/(dt)(3.5t^2+41t+9) ?

The d/(dt)(3.5t^2+41t+9) is 7t+41

What is the first d/(dt)(3.5t^2+41t+9) ?

The first d/(dt)(3.5t^2+41t+9) is 7t+41

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\frac{d}{d t} (3.5 t^{2} + 41 t + 9)

7 t + 41

Solution steps

\frac{d}{d t} (3.5 t^{2} + 41 t + 9)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (3.5 t^{2}) + \frac{d}{d t} (41 t) + \frac{d}{d t} (9)

\frac{d}{d t} (3.5 t^{2}) = 7 t

\frac{d}{d t} (41 t) = 41

\frac{d}{d t} (9) = 0

= 7 t + 41 + 0

Simplify

= 7 t + 41

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

2 tan (x) - 7 sec (x)

\frac{( 2 x ^{2} - 4 )}{( 3 x ^{2} + 7 )}

x = \frac{1}{t}

\frac{t ^{3}}{6}

What is the d/(dt)(3.5t^2+41t+9) ?
The d/(dt)(3.5t^2+41t+9) is 7t+41
What is the first d/(dt)(3.5t^2+41t+9) ?
The first d/(dt)(3.5t^2+41t+9) is 7t+41