91Ó°ÊÓ

The chain rule is a fundamental concept in calculus, particularly when dealing with composite functions—that is, functions nested within each other. To differentiate a composite function, we need to take the derivative of the outer function and multiply it by the derivative of the inner function. In simpler terms, if you have a function of the form \(y = g(f(x))\), the chain rule helps you find the derivative \(\frac{dy}{dx}\) by computing \(g'(f(x)) \cdot f'(x)\). For our original exercise, we're tasked with finding the derivative where \(y = \sqrt{f(x)}\). The chain rule comes into play as follows:

• The outer function is \(g(u) = \sqrt{u}\), which becomes \(u^{1/2}\) when expressed as a power function.
• Using the derivative found via the power rule, which is \(\frac{1}{2\sqrt{u}}\), it will then multiply the derivative of the inner function \(f(x)\).

This method helps break down complex differentiation problems into more manageable steps, simplifying the process.

The power rule is a cornerstone of basic derivative rules in calculus, making it easier to differentiate functions involving powers of \(x\). The rule states that for any function of the form \(f(x) = x^n\), its derivative is \(f'(x) = nx^{n-1}\). This simple process is very useful because it gives us a quick way to handle powers in functions. In our exercise, the power rule is applied to find the derivative of the outer function \(g(u) = u^{1/2}\). Here’s a step-by-step breakdown:

• First, express \(\sqrt{u}\) as a power, i.e., \(u^{1/2}\).
• Apply the power rule: bring down the exponent \(\frac{1}{2}\) and reduce the exponent by 1, leading to \(\frac{1}{2}u^{-1/2}\).
• Simplify the expression, resulting in \(\frac{1}{2\sqrt{u}}\).

The power rule provides a direct and efficient way to find derivatives for functions involving powers, making it a handy tool for more complex differentiation tasks.

A function being differentiable means it has a derivative at all points in its domain. Specifically, a differentiable function is smooth, with no sharp corners or cusps at any point within its interval. This is essential to calculus because differentiability implies continuity, meaning the function doesn't have any jumps or breaks.In our given function \(y=\sqrt{f(x)}\), we're told that \(f(x)\) is differentiable and nonnegative. This context assures us the derivative can be calculated smoothly:

• No points of discontinuity—ensures the behavior of \(f(x)\) is predictable.
• Nonnegative, which is crucial because the square root function only handles zero or positive values without resulting in complex numbers.

By knowing \(f(x)\) is differentiable, it guarantees that \(f'(x)\), the derivative needed for applying the chain rule, exists. This underlying differentiability is a necessary condition for using the chain rule effectively.