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The product rule is an essential technique in calculus used to differentiate functions that are multiplied together. When evaluating the derivative of a function, say \(y = uv\), where both \(u\) and \(v\) are functions of \(x\), the product rule states that: \[\frac{dy}{dx} = u'v + uv'\]

• \(u'\) is the derivative of \(u\) with respect to \(x\).
• \(v'\) is the derivative of \(v\) with respect to \(x\).

For the given function \(y = x \sqrt{1-x^2}\), it's clear that you have two distinct parts: \(u = x\) and \(v = \sqrt{1-x^2}\).
By applying the product rule, you take the derivative of \(x\) as \(1\) and then separately find the derivative of \(\sqrt{1-x^2}\). Finally, apply the rule to get the derivative of the entire function. Seeing how each part of the function is differentiated separately before combining them provides a clear path to solving more complex derivatives.

The chain rule helps us differentiate composite functions, where one function is nested inside another. It is foundational for solving derivatives where functions have an inner and outer layer.
For a function \(v = \sqrt{1-x^2}\), you can rewrite it as \(v = (1-x^2)^{1/2}\), where the outer function is the square root function and the inner function is \(1-x^2\).When using the chain rule, you do the following steps:

• Identify the outer function \(v = w^{1/2}\) and the inner function \(w = 1-x^2\).
• Differentiate the outer function with respect to the inner function. This gives \(\frac{1}{2}w^{-1/2} \).
• Differentiate the inner function \(w\) with respect to \(x\), resulting in \(-2x\).

Combine these results to get the derivative of the composite function: \[\frac{dv}{dx} = \frac{1}{2}(1-x^2)^{-1/2} \cdot (-2x) = \frac{-x}{\sqrt{1-x^2}}\].
By breaking the problem into manageable steps, you can clearly see how the chain rule is applied to reach the final derivative.

Differentiation is a central concept in calculus and describes the process of finding the derivative of a function. This process involves calculating how a function changes as its input changes.
When you apply differentiation, you are seeking a function that describes this rate of change.For the given problem \(y = x \sqrt{1-x^2}\), differentiation was employed by using both the product rule and the chain rule:

• The product rule helped to manage the multiplication of two functions: \(x\) and \(\sqrt{1-x^2}\).
• The chain rule was used to deal with the composite function, \(\sqrt{1-x^2}\).

After differentiating each part, these rules allow you to combine individual derivatives into a clean, final expression for the derivative:\[\frac{dy}{dx} = \frac{1 - 2x^2}{\sqrt{1-x^2}}\].
This shows how differentiation intertwines these fundamental rules to analyze and solve for the rate of change within complex, intertwined functions.