#### To determine

**To find:** The derivative of the function f(t).

#### Answer

The derivative of the function f(t)=2t3−3t2−4t is 6t2−6t−4.

#### Explanation

**Given:**

The function, f(t)=2t3−3t2−4t.

**Formula used:**

**The Constant Multiple Rule:**

If *c* is a constant and f(t) is a differentiable function, the constant multiple rule is,

ddt[c⋅f(t)]=c⋅ddt[f(t)] (1)

**The Power Rule:**

If n is any real number, then the power rule is,

ddt(tn)=ntn−1 (2)

**The Difference Rule:**

If f(t) and g(t) are both differentiable functions, then the difference rule is,

ddt[f(t)−g(t)]=ddt(f(t))−ddt(g(t)) (3)

**Calculation:**

The derivative of f(t)=2t3−3t2−4t is f′(t) as follows.

f′(t)=ddt(f(t)) =ddt(2t3−3t2−4t)

Apply the difference rule as shown in equation (3).

f′(t)=ddt(2t3)−ddt(3t2)−ddt(4t)

Apply the constant multiple rule as shown in equation (1).

f′(t)=2ddt(t3)−3ddt(t2)−4ddt(t)

Apply the power rule as shown in equation (2)

f′(t)=2(3t3−1)−3(2t2−1)−4(1t1−1)=2(3t2)−3(2t1)−4(1t0)=6t2−6t−4

Therefore, the derivative of the function f(t)=2t3−3t2−4t is 6t2−6t−4.