Question

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

Short answer

The derivative of the function \(y=\sqrt{x}(x-2)^{2}\) is \(y' = \frac{(x-2)^{2}}{2\sqrt{x}} + 2\sqrt{x}(x-2)\). The differentiation rules used here are the Product Rule and the Chain Rule.

Step-by-step solution

Identify the functions and rules

In this exercise, the main function \(y\) is a product of two parts: \(\sqrt{x}\) and \((x-2)^{2}\). To differentiate this, we'll use the product rule. Also, the second part of the function \((x-2)^{2}\) will require the chain rule.

Apply the Product Rule

The product rule states that the derivative of a product of two functions is the derivative of the first times the second plus the first times the derivative of the second. Therefore, \(y'\) can be shown as:\(y' = \sqrt{x}'(x-2)^{2}+ \sqrt{x}(x-2)^{2}'\)

Differentiate \(\sqrt{x}\)

The derivative of \(\sqrt{x}\) is \(\frac{1}{2\sqrt{x}}\). Therefore we can update the equation:\(y' = \frac{1}{2\sqrt{x}}(x-2)^{2}+ \sqrt{x}(x-2)^{2}'\)

Apply the Chain Rule on \((x-2)^{2}\)

The chain rule tells us how to differentiate a composition of functions. It states that the derivative of a composition of functions is the derivative of the outer function times the derivative of the inner function. Applying it to \((x-2)^{2}\), we get: \((x-2)^{2}' = 2(x-2)\times1 = 2(x-2)\). We can now insert this into the equation:\(y' = \frac{1}{2\sqrt{x}}(x-2)^{2}+ \sqrt{x}*2(x-2)\)

Simplify the Expression

Let's simplify this. It yields:\(y' = \frac{(x-2)^{2}}{2\sqrt{x}} + 2\sqrt{x}(x-2)\)