Problem 61 - Calculus - AP Edition

Question

True-False Determine whether the statement is true or false. Explain your answer. $$\text { If } y=f(x), \text { then } \frac{d}{d x}[\sqrt{y}]=\sqrt{f^{\prime}(x)}$$

Step-by-Step Solution

Answer

The statement is false.

Step 1: Understanding the Given Statement

The statement claims that the derivative of the square root of a function \(y = f(x)\) with respect to \(x\) is equal to the square root of the derivative of the function \(f(x)\). This can be represented by \(\frac{d}{dx}[\sqrt{y}] = \sqrt{f'(x)}\).

Step 2: Differentiate \(\sqrt{y}\) Using Chain Rule

The chain rule is used to differentiate composite functions. For \(y=f(x)\), we apply the chain rule to \(\sqrt{y}\): \(\frac{d}{dx}[\sqrt{y}] = \frac{d}{dy}[\sqrt{y}] \cdot \frac{dy}{dx}\). Given \(\sqrt{y} = y^{1/2}\), we differentiate: \(\frac{d}{dy}[y^{1/2}] = \frac{1}{2}y^{-1/2}\). Then multiply by \(\frac{dy}{dx} = f'(x)\): \(\frac{1}{2}y^{-1/2} \cdot f'(x)\).

Step 3: Simplify the Expression

Substitute \(y = f(x)\) back into the expression: \(\frac{1}{2}(f(x))^{-1/2} \cdot f'(x)\). Simplifying further gives \(\frac{f'(x)}{2\sqrt{f(x)}}\).

Step 4: Compare the Derivative Expression to the Given Statement

The computed derivative \(\frac{f'(x)}{2\sqrt{f(x)}}\) is not equal to \(\sqrt{f'(x)}\) as proposed in the statement. The expression \(\sqrt{f'(x)}\) lacks the \( \frac{1}{2}\) factor and the division by \(\sqrt{f(x)}\), indicating a significant difference.

Key Concepts

Derivative of Composite Function

When dealing with the derivative of composite functions, an essential concept in calculus, we often employ the chain rule to differentiate such functions efficiently. A composite function can generally be understood as a function within a function, such as \(y = f(g(x))\).
To find the derivative of a composite function, we apply the chain rule, which says:

• Differentiation of the outer function with respect to its inner function.
• Multiplying this by the derivative of the inner function.

Applying this to \(y = f(x)\) and \(\sqrt{y}\):

• We first find the derivative of \(\sqrt{y}\) with respect to \(y\), which is \(\frac{1}{2} y^{-1/2}\).
• Then, we multiply it by the derivative of \(y = f(x)\) with respect to \(x\), which is \(f'(x)\).

Thus, the chain rule allows us to successfully find derivatives of more complex expressions involving multiple functions together.

Square Root Differentiation

In calculus, differentiating a square root function is straightforward if we remember it's just a special case of exponential functions. The square root of a variable \(y\) is expressed as \(y^{1/2}\). To find its derivative, we use basic power differentiation rules.
To differentiate \(y^{1/2}\):

• We apply the power rule, which gives us \(\frac{1}{2} y^{-1/2}\).

This fractional exponent form simplifies differentiation whilst allowing for continuation in more complex operations when \(y\) itself is a function.

Through this understanding, we can handle more elaborate calculus problems involving square roots systematically.

Calculus Problem Solving

Solving calculus problems involves understanding and applying various differentiation rules effectively to arrive at correct solutions. The problem given provides an interesting example where careful application of the chain rule and differentiation techniques are crucial.

In our exercise, we needed to differentiate \(\sqrt{y}\), where \(y=f(x)\). Following through with correct application:

• Application of the chain rule helped to differentiate \(\sqrt{y}\).
• Revised understanding of square root differentiation ensured we applied the principles correctly.

These thorough steps provide clarity that directly assessing complicated statements without checking can lead to incorrect conclusions, such as assuming the original assertion: \(\frac{d}{dx} [\sqrt{y}] = \sqrt{f'(x)}\) is true.
By systematically breaking down and analyzing each component, calculus problem solving becomes a manageable and methodical process.