What is the derivative of f(t)=3+8t-t^2 ?

The derivative of f(t)=3+8t-t^2 is -2t+8

What is the first derivative of f(t)=3+8t-t^2 ?

The first derivative of f(t)=3+8t-t^2 is -2t+8

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derivative of

f (t) = 3 + 8 t - t^{2}

- 2 t + 8

Solution steps

\frac{d}{d t} (3 + 8 t - t^{2})

Apply the Sum/Difference Rule:

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

= \frac{d}{d t} (3) + \frac{d}{d t} (8 t) - \frac{d}{d t} (t^{2})

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

\frac{d}{d t} (8 t) = 8

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

= 0 + 8 - 2 t

Simplify

= - 2 t + 8

x^{2} cos (\frac{1}{x})

y = x^{2} + 5

20 q - 2 q^{2}

s + 1

f (x) = (x^{6} + 4)^{5}

What is the derivative of f(t)=3+8t-t^2 ?
The derivative of f(t)=3+8t-t^2 is -2t+8
What is the first derivative of f(t)=3+8t-t^2 ?
The first derivative of f(t)=3+8t-t^2 is -2t+8