Problem 61 - College Algebra and Calculus: An Applied Approach

Question

Find the derivative of the function. State which differentiation rule(s) you used to find the derivative. $$ y=x \sqrt{2 x+3} $$

Step-by-Step Solution

Answer

The derivative of the function \(y=x \sqrt{2x+3}\) is \(y'(x) = \sqrt{2x+3} + \frac{x}{\sqrt{2x+3}}\). The Product Rule and the Chain Rule were used for finding the derivative.

Step 1: Identify the Functions

Given \(y = x \sqrt{2x+3}\), identify the two functions as \(f(x) = x\) and \(g(x) = \sqrt{2x+3}\)

Step 2: Apply Product Rule

Apply the Product Rule which is \((f \cdot g)'=f'g+fg'\) for \(y=x \sqrt{2x+3}\). Now calculate the derivative of \(f(x) = x\), which is \(f'(x) = 1\).

Step 3: Apply Chain Rule

The function \(g(x) = \sqrt{2x+3}\) is a composite function, so the Chain Rule will be used for differentiation. The Chain Rule is \((f(g(x)))' = f'(g(x)) \cdot g'(x)\). Rewrite the function in the form \(g(x) = (2x + 3)^{0.5}\), then apply the Chain Rule. Derive the outer function first keeping the inner function as is, then derive the inner function. Doing this gives us \(g'(x) = 0.5\cdot(2x+3)^{-0.5} \cdot 2\).

Step 4: Combine the Results

Combine the derivatives of \(f(x)\) and \(g(x)\) according to the Product Rule. Thus, \(y' = f'g+fg' = 1 \cdot \sqrt{2x+3} + x \cdot 0.5 \cdot (2x+3)^{-0.5} \cdot 2\). Simplify this to get \(y'(x) = \sqrt{2x+3} + \frac{x}{\sqrt{2x+3}}\).