What is the derivative of 3+vt-4.9t^2 ?

The derivative of 3+vt-4.9t^2 is t

What is the first derivative of 3+vt-4.9t^2 ?

The first derivative of 3+vt-4.9t^2 is t

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derivative of

3 + v t - 4.9 t^{2}

t

Solution steps

\frac{d}{d v} (3 + v t - 4.9 t^{2})

Treat

t

as a constant

Apply the Sum/Difference Rule:

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

= \frac{d}{d v} (3) + \frac{d}{d v} (v t) - \frac{d}{d v} (4.9 t^{2})

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

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

\frac{d}{d v} (4.9 t^{2}) = 0

= 0 + t - 0

Simplify

= t

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What is the derivative of 3+vt-4.9t^2 ?
The derivative of 3+vt-4.9t^2 is t
What is the first derivative of 3+vt-4.9t^2 ?
The first derivative of 3+vt-4.9t^2 is t