Problem 33 - College Algebra and Calculus: An Applied Approach

Question

Use the limit definition to find the derivative of the function. $$ h(t)=\sqrt{t-1} $$

Step-by-Step Solution

Answer

The derivative of the function \( h(t) = \sqrt{t - 1} \) is \( h'(t) = \frac{1}{2(t - 1)^{1/2}} \).

Step 1: Rewrite the function

Firstly, express the function in a more convenient form for differentiation. Here, \( h(t) = \sqrt{t - 1} \) can be rewritten as \( h(t) = (t - 1)^{1/2} \). It is easier to differentiate in this form.

Step 2: Apply the limit definition

The limit definition of a derivative is \( h'(t) = \lim_{s\to t} \frac{h(s) - h(t)}{s - t} \). Apply this definition with the function \( h(t) = (t - 1)^{1/2} \). You get \( h'(t) = \lim_{s\to t} \frac{(s - 1)^{1/2} - (t - 1)^{1/2}}{s - t} \). Now the task is to simplify the right side of the equation.

Step 3: Simplify the expression

Multiplying the numerator and denominator by the conjugate of the numerator will get rid of the fraction. The conjugate of the numerator is \( (s + 1)^{1/2} + (t - 1)^{1/2} \), and after multiplication, you get: \( h'(t) = \lim_{s\to t} \frac{(s - t)}{(s - t)[(s - 1)^{1/2} + (t - 1)^{1/2}]} = \frac{1}{2(t - 1)^{1/2}} \cdot (1 - 0) = \frac{1}{2(t - 1)^{1/2}} \)

Step 4: Evaluating the limit

Now, simply evaluate the limit as \( s \to t \) which will lead to the final derivative.