Question

True or False? In Exercises 137-139, determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. If \(y=(1-x)^{1 / 2},\) then \(y^{\prime}=\frac{1}{2}(1-x)^{-1 / 2}\)

Short answer

False, if \(y=(1-x)^{1 / 2}\), then \(y' = - \frac{1}{2}(1-x)^{-1 / 2}\), not \( \frac{1}{2}(1-x)^{-1 / 2}\).

Step-by-step solution

Identify the Function and its components

The given function is \(y=(1-x)^{1 / 2}\). Here, the inner function \(u\) is \((1-x)\) and the outer function \(v\) is \(u^{1 / 2}\). This is a composition of functions, so we're going to need the chain rule to differentiate it.

Differentiate Using Chain Rule

According to chain rule, \( \frac{dy}{dx} = \frac{dv}{du} \cdot \frac{du}{dx}\). We have \(\frac{du}{dx}= -1\) and \(\frac{dv}{du}= \frac{1}{2}u^{-1 / 2} = \frac{1}{2}(1-x)^{-1 / 2}\). So applying chain rule, we obtain \(y' = \frac{dv}{du} \cdot \frac{du}{dx} = - \frac{1}{2}(1-x)^{-1 / 2}\)

Compare the result

The derivative we found, \(y' = - \frac{1}{2}(1-x)^{-1 / 2}\), is not the same as the one given in the problem, which is \( \frac{1}{2}(1-x)^{-1 / 2}\). Therefore, the statement is False.

Explanation of False Statement

If \(y=(1-x)^{1 / 2}\), then \(y' = - \frac{1}{2}(1-x)^{-1 / 2}\), not \( \frac{1}{2}(1-x)^{-1 / 2}\). The negative sign comes from differentiating the inner function (1-x), which yields -1.