What is the d/(dt)(3t^3-2t^2+t-5) ?

The d/(dt)(3t^3-2t^2+t-5) is 9t^2-4t+1

What is the first d/(dt)(3t^3-2t^2+t-5) ?

The first d/(dt)(3t^3-2t^2+t-5) is 9t^2-4t+1

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\frac{d}{d t} (3 t^{3} - 2 t^{2} + t - 5)

9 t^{2} - 4 t + 1

Solution steps

\frac{d}{d t} (3 t^{3} - 2 t^{2} + t - 5)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (3 t^{3}) - \frac{d}{d t} (2 t^{2}) + \frac{d t}{d t} - \frac{d}{d t} (5)

\frac{d}{d t} (3 t^{3}) = 9 t^{2}

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

\frac{d t}{d t} = 1

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

= 9 t^{2} - 4 t + 1 - 0

Simplify

= 9 t^{2} - 4 t + 1

What is the d/(dt)(3t^3-2t^2+t-5) ?
The d/(dt)(3t^3-2t^2+t-5) is 9t^2-4t+1
What is the first d/(dt)(3t^3-2t^2+t-5) ?
The first d/(dt)(3t^3-2t^2+t-5) is 9t^2-4t+1