What is the d/(dt)((x)(-2x^2+16x)) ?

The d/(dt)((x)(-2x^2+16x)) is -6t^2+32t

What is the first d/(dt)((x)(-2x^2+16x)) ?

The first d/(dt)((x)(-2x^2+16x)) is -6t^2+32t

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\frac{d}{d t} ((t) (- 2 t^{2} + 16 t))

- 6 t^{2} + 32 t

Solution steps

\frac{d}{d t} ((t) (- 2 t^{2} + 16 t))

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = t, g = - 2 t^{2} + 16 t

= \frac{d t}{d t} (- 2 t^{2} + 16 t) + \frac{d}{d t} (- 2 t^{2} + 16 t) t

\frac{d t}{d t} = 1

\frac{d}{d t} (- 2 t^{2} + 16 t) = - 4 t + 16

= 1 \cdot (- 2 t^{2} + 16 t) + (- 4 t + 16) t

Simplify

1 \cdot (- 2 t^{2} + 16 t) + (- 4 t + 16) t : - 6 t^{2} + 32 t

= - 6 t^{2} + 32 t

d x (6 x^{3} + 2 x + 1)

\frac{s - 1}{s ^{2} ( s - 2 ) ( s + 1 )}

a r g [(4 - 3 i) (- 1 + 2 i)]

, 7 (x + h)^{2} - 4 (x + h) + 7

What is the d/(dt)((x)(-2x^2+16x)) ?
The d/(dt)((x)(-2x^2+16x)) is -6t^2+32t
What is the first d/(dt)((x)(-2x^2+16x)) ?
The first d/(dt)((x)(-2x^2+16x)) is -6t^2+32t