What is the derivative of P(t)=0.45t^2-12t+105 ?

The derivative of P(t)=0.45t^2-12t+105 is 0.9t-12

What is the first derivative of P(t)=0.45t^2-12t+105 ?

The first derivative of P(t)=0.45t^2-12t+105 is 0.9t-12

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derivative of

P (t) = 0.45 t^{2} - 12 t + 105

0.9 t - 12

Solution steps

\frac{d}{d t} (0.45 t^{2} - 12 t + 105)

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (0.45 t^{2}) - \frac{d}{d t} (12 t) + \frac{d}{d t} (105)

\frac{d}{d t} (0.45 t^{2}) = 0.9 t

\frac{d}{d t} (12 t) = 12

\frac{d}{d t} (105) = 0

= 0.9 t - 12 + 0

Simplify

= 0.9 t - 12

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What is the derivative of P(t)=0.45t^2-12t+105 ?
The derivative of P(t)=0.45t^2-12t+105 is 0.9t-12
What is the first derivative of P(t)=0.45t^2-12t+105 ?
The first derivative of P(t)=0.45t^2-12t+105 is 0.9t-12