#### To determine

**To find:** The derivative of the function g(t)=1t and state the domain of the function and its derivative.

#### Answer

The derivative of the function g(t) is −12t32_.

The domain of the function g(t) is (0,∞)

The domain of g′(t) is (0,∞).

#### Explanation

**Formula used:**

The derivative of a function *f ,* denoted by f′(x), is

f′(x)=limh→0f(x+h)−f(x)h (1)

**Difference of squares formula:** (a2−b2)=(a+b)(a−b)

**Calculation:**

Obtain the derivative of the function g(t).

Use the equation (1) to compute g′(t),

g′(t)=limh→0g(t+h)−g(t)h=limh→01t+h−1th=limh→0t−t+htt+hh=limh→0(t−t+h)htt+h

Multiply both the numerator and the denominator by the conjugate of the numerator,

g′(t)=limh→0(t−t+h)htt+h×(t+t+h)(t+t+h)=limh→0(t−t+h)(t+t+h)htt+h(t+t+h)

Apply the difference of square formula,

g′(t)=limh→0(t)2−(t+h)2h(tt+ht+tt+ht+h)=limh→0t−(t+h)h((t)2t+h+t(t+h)2)=limh→0t−t−hh(tt+h+t(t+h))=limh→0−hh(tt+h+t(t+h))

Since the limit *h* approaches zero but not equal to zero, cancel the common term *h* from both the numerator and the denominator,

g′(t)=limh→0−1(tt+h+t(t+h))=−1tt+(0)+t(t+(0))=−1tt+tt=−12tt

=−12t32

Thus, the derivative of the function g(t) is −12t32_.

The function is defined for every real numbers except (−∞,0].

Therefore, the domain of the function g(t)=1t is (0,∞).

The domain of the derivative g′(t) is {t∈ℝ| g′(t) exists}.

Since the derivative g′(t)=−12t32 exists for all real numbers except (−∞,0].

Therefore, the domain of g′(t) is (0,∞).