#### To determine

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

#### Answer

The derivative of the function f(t)=1.4t5−2.5t2+6.7 is 7t4−5t.

#### Explanation

**Given:**

The function, f(t)=1.4t5−2.5t2+6.7.

**Formula used:**

**Derivative of a Constant Function:**

If *c* is a constant function, then ddt(c)=0 (1)

**The Constant Multiple Rule:**

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

ddt[c⋅f(t)]=c⋅ddtf(t) (2)

**The Power Rule:**

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

ddt(tn)=ntn−1 (3)

**The Sum Rule:**

If f(t) and g(t) both are differentiable function, then the sum rule is,

ddt[f(t)+g(t)]=ddt[f(t)]+ddt[g(t)] (4)

**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)] (5)

**Calculation:**

The derivative of f(t)=1.4t5−2.5t2+6.7 is f′(t) as follows.

f′(t)=ddt[f(t)] =ddt(1.4t5−2.5t2+6.7)

Apply the sum rule as shown in equation (4).

f′(t)=ddt(1.4t5−2.5t2)+ddt(6.7)

Apply the difference rule as shown in equation (5).

f′(t)=1.4ddt(t5)−2.5ddt(t2)+ddt(6.7)

Apply the constant multiple rule as shown in equation (2).

f′(t)=1.4ddt(t5)−2.5ddt(t2)+ddt(6.7)

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

f′(t)=1.4(5t5−1)−2.5(2t2−1)+ddt(6.7)

Apply the rule of derivative of constant function.

f′(t)=1.4(5t5−1)−2.5(2t2−1)+0 =1.4(5t4)−2.5(2t1)=7t4−5t

Therefore, the derivative of the function f(t)=1.4t5−2.5t2+6.7 is 7t4−5t_.