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

We save the definition:

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

Let's differentiate:

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

ⓘ A constant factor stays in front of the derivative, (k*u)' = k*u':

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

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

■The derivative of a sum is the sum of the derivatives:

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

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

0.1 \cdot (0 + \frac{d}{d t} (- 40 t) + \frac{d}{d t} (t^{2}))

ⓘ Adding 0 changes nothing:

0 + \frac{d}{d t} (- 40 t)

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

■A constant factor stays in front of the derivative, (k*u)' = k*u':

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

-40

■The derivative of x^n is n*x^(n-1):

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

2 t

0.1 \cdot (- 40 + 2 t)

0.1 \cdot (2 t - 40)

⇥Let's factor in

ℝ

(2 t - 40)

⇥We factor out

2

, the greatest common divisor of the coefficients:

2 t - 40

2 \cdot (t - 20)

⇥Factor identified:

t - 20

0.1 \cdot 2 \cdot (t - 20)

ⅰ

0.1 \cdot 2

0.1 + 0.1

0.2

▶Derivative:

0.2 \cdot (t - 20)