Question

Determine whether the statement is true or false. If it is false, explain why or give an example that shows it is false. $$ \text { If } y=(1-x)^{1 / 2} \text { , then } y^{\prime}=\frac{1}{2}(1-x)^{-1 / 2} \text { . } $$

Short answer

The original statement is false. The correct derivative should be \(-1/2 (1-x)^{-1/2}\).

Step-by-step solution

Analyze the Given Function

The function given is \(y = (1-x)^{1/2}\). It's essentially a square root function where the argument is \(1-x\). The function is differentiated with respect to \(x\).

Calculate The Derivative Using Chain Rule

First we need to apply the chain rule for differentiation, which states that the derivative of a composite function is the derivative of the outside function times the derivative of the inside function. In our case, the outside function is \(u^{1/2}\) and the inside function is \(1-x\). So the chain rule gives us: \((u^{1/2})' * (1-x)'\), which evaluates to: \(\frac{1}{2}(1-x)^{-1/2}*-1\).

Evaluate the Result and Compare with Given Derivative

After applying the chain rule, we get \(-1/2 (1-x)^{-1/2}\). The negative sign indicates that the slope will decrease as \(x\) increases. Comparing this with the given derivative \(\frac{1}{2}(1-x)^{-1/2}\), it's clear that they are not the same. Therefore, the original statement is false.

Key concepts

Differentiation

Differentiation is a foundational concept in calculus that involves calculating the rate at which a function changes. This is often represented as the derivative of the function.

Think of differentiation as a tool that helps us understand how a small change in one variable can affect another variable linked to it. It's commonly used to find slopes of curves or the instantaneous rate of change at a point.

In mathematical terms, if you have a function, say, \( f(x) \), its derivative is written as \( f'(x) \). This tells you how much \( f(x) \) changes as \( x \) changes just a tiny bit. To calculate it, you need to know certain rules, like the power rule, product rule, quotient rule, and chain rule.

In our exercise, we applied differentiation to a square root function. This involved using the chain rule, a special technique to handle more complex functions. Differentiation isn't just for math whizzes; it's a practical tool for many real-world problems involving rates of change.

Composite Function

Composite functions are fascinating! They occur when one function is nested inside another. Imagine having two functions, \( f(x) \) and \( g(x) \). If you compose them, you create a new function, like \( f(g(x)) \).

In our exercise, the function \( y = (1-x)^{1/2} \) is a great example. It's a composite function where the inside function is \( the,\ 1-x \) and the outside function is \( u^{1/2} \). This kind of function requires special attention when differentiating, which brings us to the chain rule.

Composite functions are everywhere. From physics to finance, they help model scenarios where an input must pass through multiple stages before producing an output. Understanding how they work is crucial for tackling complex differentiation tasks.

Derivative Calculation

Calculating derivatives may seem challenging at first, but it gets easier once you grasp some rules and techniques. When you have a composite function like \( y = (1-x)^{1/2} \), the derivative calculation involves some key steps using the chain rule.

Here's our step-by-step approach:

• Identify both the outside and inside functions. For \( y = (1-x)^{1/2} \), the outside is \( u^{1/2} \) and the inside is \( 1-x \).
• Apply the chain rule, which says take the derivative of the outside part (\( \frac{1}{2}u^{-1/2} \)) and multiply it by the derivative of the inside part (\( -1 \)).
• Combine them to get the result, \( -1/2 (1-x)^{-1/2} \). This shows the correct derivative.
• Notice the difference between this and what was incorrectly given as \( \frac{1}{2}(1-x)^{-1/2} \). The missing negative sign changes the meaning significantly and verifies the original statement as false.

The derivative calculation process is powerful. With practice, it can become a very intuitive, almost second-nature task when dealing with functions in calculus.