What is the derivative of (4t^2-t)(t^3-8t^2+12) ?

The derivative of (4t^2-t)(t^3-8t^2+12) is 20t^4-132t^3+24t^2+96t-12

What is the first derivative of (4t^2-t)(t^3-8t^2+12) ?

The first derivative of (4t^2-t)(t^3-8t^2+12) is 20t^4-132t^3+24t^2+96t-12

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derivative of

(4 t^{2} - t) (t^{3} - 8 t^{2} + 12)

20 t^{4} - 132 t^{3} + 24 t^{2} + 96 t - 12

Solution steps

\frac{d}{d t} ((4 t^{2} - t) (t^{3} - 8 t^{2} + 12))

Apply the Product Rule:

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

f = 4 t^{2} - t, g = t^{3} - 8 t^{2} + 12

= \frac{d}{d t} (4 t^{2} - t) (t^{3} - 8 t^{2} + 12) + \frac{d}{d t} (t^{3} - 8 t^{2} + 12) (4 t^{2} - t)

\frac{d}{d t} (4 t^{2} - t) = 8 t - 1

\frac{d}{d t} (t^{3} - 8 t^{2} + 12) = 3 t^{2} - 16 t

= (8 t - 1) (t^{3} - 8 t^{2} + 12) + (3 t^{2} - 16 t) (4 t^{2} - t)

Simplify

(8 t - 1) (t^{3} - 8 t^{2} + 12) + (3 t^{2} - 16 t) (4 t^{2} - t) : 20 t^{4} - 132 t^{3} + 24 t^{2} + 96 t - 12

= 20 t^{4} - 132 t^{3} + 24 t^{2} + 96 t - 12

y = sin (x), y = 2 sin^{2} (x), x = 0, x = \frac{π}{2}

x \to π - lim (\frac{x}{cos ( x ) + 1})

y^{^{''}} + 4 y^{^{'}} + 3 y = 0

x \to 0 - lim (4 + \frac{6}{x ^{2}})

f (x) = x^{2}, at x = - 1

What is the derivative of (4t^2-t)(t^3-8t^2+12) ?
The derivative of (4t^2-t)(t^3-8t^2+12) is 20t^4-132t^3+24t^2+96t-12
What is the first derivative of (4t^2-t)(t^3-8t^2+12) ?
The first derivative of (4t^2-t)(t^3-8t^2+12) is 20t^4-132t^3+24t^2+96t-12