Question

Prove: \(\quad D_{z} \sqrt{1-x}=-\frac{1}{2 \sqrt{1-x}}\)

Short answer

Question: Prove that the derivative of the function \(\sqrt{1-x}\) with respect to \(x\) is equal to \(-\frac{1}{2\sqrt{1-x}}\). Answer: To prove that the derivative of the function \(\sqrt{1-x}\) with respect to \(x\) is equal to \(-\frac{1}{2\sqrt{1-x}}\), we followed these steps: 1) Identify the function to differentiate: \(f(x) = \sqrt{1-x}\) 2) Rewrite the function: \(f(x) = (1-x)^{\frac{1}{2}}\) 3) Apply the power rule for differentiation: \(\frac{d}{dx}\left((1-x)^{\frac{1}{2}}\right) = \frac{1}{2}(1-x)^{-\frac{1}{2}}\cdot(-1)\) 4) Simplify the result: \(\frac{d}{dx}\left((1-x)^{\frac{1}{2}}\right)= -\frac{1}{2}(1-x)^{-\frac{1}{2}}\) 5) Rewrite the result in the original form: \(D_{z}\sqrt{1-x}=-\frac{1}{2\sqrt{1-x}}\). Thus, the proof is complete: \(D_{z}\sqrt{1-x}=-\frac{1}{2\sqrt{1-x}}\).

Step-by-step solution

Identify the function to differentiate

The problem asks for the derivation of the function \(f(x) = \sqrt{1-x}\).

Rewrite the function

Rewrite the function as \(f(x) = (1-x)^{\frac{1}{2}}\)

Apply the power rule for differentiation

Using the power rule for differentiation, we have: \(\frac{d}{dx}\left((1-x)^{\frac{1}{2}}\right) = \frac{1}{2}(1-x)^{-\frac{1}{2}}\cdot(-1)\)

Simplify the result

Now, simplify the result: \(\frac{d}{dx}\left((1-x)^{\frac{1}{2}}\right)= -\frac{1}{2}(1-x)^{-\frac{1}{2}}\)

Rewrite the result in original form

Rewrite the result in the original form: \(D_{z}\sqrt{1-x}=-\frac{1}{2\sqrt{1-x}}\). Thus, the proof is complete: \(D_{z}\sqrt{1-x}=-\frac{1}{2\sqrt{1-x}}\).

Key concepts

Power Rule for Differentiation

Differentiation is a fundamental concept in calculus, essential for understanding change and motion. One of the most powerful and frequently used techniques in finding derivatives is known as the power rule. The power rule provides a quick and straightforward method for differentiating functions of the form
\( f(x) = x^n \)
where \( n \) is any real number. According to this rule, the derivative of \( f(x) \) with respect to \( x \) is:
\( f'(x) = nx^{n-1} \).

In our exercise, we encounter the function \( f(x) = (1-x)^{\frac{1}{2}} \). To apply the power rule effectively, we must recognize that the exponent \( \frac{1}{2} \) denotes a square root and rewrite the function accordingly. Once rewritten, the power rule is used to find the derivative, resulting in a multiplied factor that reflects the exponent while reducing the original exponent by one. Furthermore, we must consider the negative sign inside the bracket, introducing a multiplication by \( -1 \) during differentiation— a critical step for obtaining correct solutions.

Derivative of a Function

The derivative of a function at a certain point measures how the function's output value changes as the input value changes. In essence, it describes the slope of the function at any point on its curve. To determine the derivative of a function, we use rules of differentiation like the power rule mentioned above.

In the context of our textbook exercise, we observe that the derivative is not just some numeric value but rather a new function that gives the slope of \( \sqrt{1-x} \) at any point \( x \). This concept is particularly powerful as it allows us to analyze the behavior of functions beyond the mere calculation of tangent slopes at specific points—it leads to understanding instantaneous rates of change and forms the foundation for further studies in calculus, such as integration and solving differential equations.

Calculus Proofs

Calculus proofs are logical arguments that verify the truths of mathematical statements, including formulas and rules of differentiation. They are essential because they not only confirm that our solutions are correct but also deepen our understanding of why certain mathematical relationships exist. A proof, like the one shown in our example, requires a clear statement of what is to be proven, followed by a logical sequence of steps that lead to the conclusion.

During the proof process, we meticulously apply known rules and properties—such as the power rule—to derive the result. As seen in the solution of the given exercise, we began by identifying the function, then rewriting it to a more suitable form for differentiation. After applying differentiation rules and simplifying, we arrived at the desired result. The proof concludes by restating the finding in its original context, which provides the necessary closure to ascertain that the differentiation has been performed correctly. Proofs exemplify rigorous reasoning and are a cornerstone of advanced mathematical study.