What is the derivative of v(t)=30-t^2 ?

The derivative of v(t)=30-t^2 is -2t

What is the first derivative of v(t)=30-t^2 ?

The first derivative of v(t)=30-t^2 is -2t

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derivative of

v (t) = 30 - t^{2}

- 2 t

Solution steps

\frac{d}{d t} (30 - t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (30) - \frac{d}{d t} (t^{2})

\frac{d}{d t} (30) = 0

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

= 0 - 2 t

Simplify

= - 2 t

\frac{\partial}{\partial x} (sin (x^{3}) + 3 x y)

t^{2} cos (4 t)

y = x^{2} - 8 x, at x = - 9

\int ye d y

\int \frac{1}{x ^{2} - 2 x + 2} d x

What is the derivative of v(t)=30-t^2 ?
The derivative of v(t)=30-t^2 is -2t
What is the first derivative of v(t)=30-t^2 ?
The first derivative of v(t)=30-t^2 is -2t