What is the derivative of-19.6t^2+9t+3 ?

The derivative of-19.6t^2+9t+3 is -39.2t+9

What is the first derivative of-19.6t^2+9t+3 ?

The first derivative of-19.6t^2+9t+3 is -39.2t+9

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derivative of

- 19.6 t^{2} + 9 t + 3

- 39.2 t + 9

Solution steps

\frac{d}{d t} (- 19.6 t^{2} + 9 t + 3)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (19.6 t^{2}) + \frac{d}{d t} (9 t) + \frac{d}{d t} (3)

\frac{d}{d t} (19.6 t^{2}) = 39.2 t

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

\frac{d}{d t} (3) = 0

= - 39.2 t + 9 + 0

Simplify

= - 39.2 t + 9

\frac{d}{d x} (40320)

f (x) = \frac{x ^{3}}{6}

\int_{3}^{6} \frac{x ^{2} - 9}{x} d x

x \to 0 lim (\frac{1}{x - 1})

\int (x - 1) (x^{2} - 2 x)^{2} d x

What is the derivative of-19.6t^2+9t+3 ?
The derivative of-19.6t^2+9t+3 is -39.2t+9
What is the first derivative of-19.6t^2+9t+3 ?
The first derivative of-19.6t^2+9t+3 is -39.2t+9