What is the derivative of y=-2t^2+3t-6 ?

The derivative of y=-2t^2+3t-6 is -4t+3

What is the first derivative of y=-2t^2+3t-6 ?

The first derivative of y=-2t^2+3t-6 is -4t+3

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derivative of

y = - 2 t^{2} + 3 t - 6

- 4 t + 3

Solution steps

\frac{d}{d t} (- 2 t^{2} + 3 t - 6)

Apply the Sum/Difference Rule:

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

= - \frac{d}{d t} (2 t^{2}) + \frac{d}{d t} (3 t) - \frac{d}{d t} (6)

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

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

\frac{d}{d t} (6) = 0

= - 4 t + 3 - 0

Simplify

= - 4 t + 3

\frac{\partial}{\partial s} (r + s)

\int \frac{tan ( x )}{sec ( x )} d x

\frac{d}{d x} (\frac{2 x}{1 - tan ( x )})

y = 0, y = - x + 5, y = x + 1

- \frac{s - 2}{2 ( s ^{2} - 2 s + 2 )}

What is the derivative of y=-2t^2+3t-6 ?
The derivative of y=-2t^2+3t-6 is -4t+3
What is the first derivative of y=-2t^2+3t-6 ?
The first derivative of y=-2t^2+3t-6 is -4t+3