Problem 34 - College Algebra and Calculus: An Applied Approach

Question

Use the limit definition to find the derivative of the function. $$ f(x)=\sqrt{x+2} $$

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = \sqrt{x+2}\) is \(f'(x) =\frac{1}{2\sqrt{x+2}}\).

Step 1: Function Substitution

Replace \(f(x + h)\) and \(f(x)\) in the limit definition of the derivative, with \(f(x) = \sqrt{x + 2}\). This gives, \( f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+2} - \sqrt{x + 2}}{h}\).

Step 2: Rationalize the numerator

Multiply the fraction by the conjugate of the numerator, without changing the value of the expression: \( f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h+2} - \sqrt{x + 2}}{h} * \frac{\sqrt{x+h+2}+ \sqrt{x+2}}{\sqrt{x+h+2}+ \sqrt{x+2}}\). Doing so, we obtain \(f'(x) = \lim_{h \to 0} \frac{(x+h+2) - (x+2)}{h*(\sqrt{x+h+2}+ \sqrt{x+2})}\). Simplifying this expression gives \(f'(x) = \lim_{h \to 0} \frac{h}{h*(\sqrt{x+h+2}+ \sqrt{x+2})}\).

Step 3: Simplify

The \(h\) in the numerator and denominator will cancel out, resulting in: \(f'(x) = \lim_{h \to 0} \frac{1}{\sqrt{x+h+2} + \sqrt{x+2}}\)

Step 4: Evaluate the Limit

As \(h\) tends to 0, the expression simplifies to: \(f'(x) =\frac{1}{2\sqrt{x+2}}\)