What is the derivative of f(t)=0.1(400-40t+t^2) ?

The derivative of f(t)=0.1(400-40t+t^2) is 0.1(2t-40)

What is the first derivative of f(t)=0.1(400-40t+t^2) ?

The first derivative of f(t)=0.1(400-40t+t^2) is 0.1(2t-40)

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derivative of

f (t) = 0.1 (400 - 40 t + t^{2})

0.1 (2 t - 40)

Solution steps

\frac{d}{d t} (0.1 (400 - 40 t + t^{2}))

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= 0.1 \frac{d}{d t} (400 - 40 t + t^{2})

Apply the Sum/Difference Rule:

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

= 0.1 (\frac{d}{d t} (400) - \frac{d}{d t} (40 t) + \frac{d}{d t} (t^{2}))

\frac{d}{d t} (400) = 0

\frac{d}{d t} (40 t) = 40

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

= 0.1 (0 - 40 + 2 t)

Simplify

= 0.1 (2 t - 40)

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What is the derivative of f(t)=0.1(400-40t+t^2) ?
The derivative of f(t)=0.1(400-40t+t^2) is 0.1(2t-40)
What is the first derivative of f(t)=0.1(400-40t+t^2) ?
The first derivative of f(t)=0.1(400-40t+t^2) is 0.1(2t-40)