91Ó°ÊÓ

When dealing with the differentiation of products of two functions, the product rule is a key tool. It states that if you have two functions, say \( u(x) \) and \( v(x) \), their derivative can be found using: \ \( \frac{d}{dx} (u v) = u \frac{d}{dx} (v) + v \frac{d}{dx} (u) \). This means, to find the derivative of the product of two functions, we differentiate each function in turn while keeping the other one constant, and then sum the results. For example, when differentiating \( x^2 \sin(3x) \), we let \( u = x^2 \) and \( v = \sin(3x) \). Here, \( u' = 2x \) and \( v' = 3 \cos(3x) \). Using the product rule, we have: \ \( \frac{d}{dx}(x^2 \sin(3x)) = 2x \sin(3x) + 3x^2 \cos(3x) \).

The chain rule is essential when differentiating composite functions. This rule states that if you have a function inside another function, say \( f(g(x)) \), the derivative is: \ \( \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \). Practically, this means we first differentiate the outer function, then multiply by the derivative of the inner function. As an example, consider \( \mathrm{e}^{-2/x} \), we let \( u = -2/x \). Then, applying the chain rule gives us: \ \( \frac{d}{dx} \mathrm{e}^{-2/x} = \mathrm{e}^{-2/x} \cdot \frac{d}{dx}(-2/x) = \mathrm{e}^{-2/x} \cdot \frac{2}{x^2} \), which simplifies to \( \frac{2 \mathrm{e}^{-2/x}}{x^2} \).

When differentiating the quotient of two functions, the quotient rule is used. It states that for functions \( u(x) \) and \( v(x) \), the derivative of \( \frac{u}{v} \) is: \ \( \frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \cdot u' - u \cdot v'}{v^2} \). Essentially, differentiate the numerator and denominator separately, multiply them appropriately, subtract, and then divide by the square of the denominator. Applying this to \( \left(\frac{x-1}{2-x}\right)^2 \), we first recognize \( f(u) = u^2 \) and \( u = \frac{x-1}{2-x} \). Using quotient and chain rules together: \ \( \frac{d}{dx} \left( \left( \frac{x-1}{2-x} \right)^2 \right) = \frac{d}{dx} \left( \left( \frac{x-1}{2-x} \right) \right) \cdot 2u = 2 \left( \frac{x-1}{2-x} \right) \cdot \frac{1}{(2-x)^2} = \frac{2(x-1)}{(2-x)^3} \).

The second derivative \( \frac{d^2y}{dx^2} \) is the derivative of the first derivative. In other words, it measures how the rate of change of a function itself changes. To find the second derivative, you might need to apply differentiation rules successively. For instance, starting with \( y = \ln(1 + \sin x) \), we found the first derivative: \ \( \frac{dy}{dx} = \frac{\cos(x)}{1 + \sin(x)} \). Now, to find the second derivative, we take the derivative of \( \frac{dy}{dx} \). Using quotient rule again: \ \( \frac{d^2y}{dx^2} = \frac{(1 + \sin x)(-\sin x) - \cos x \cdot \cos x}{(1 + \sin x)^2} \), which simplifies to \( \frac{-\sin x - \sin^2 x - \cos^2 x}{(1 + \sin x)^2} = \frac{-\sin x - 1}{(1 + \sin x)^2} \). It is crucial in geometric interpretations and applications like finding curvature.

Local maxima represent points where a function reaches a peak in a neighborhood. Mathematically, if \( y \) has a local maximum at \( x = a \), then \( \frac{dy}{dx} = 0 \) and \( \frac{d^2y}{dx^2} < 0 \) at \( x = a \). To deduce this from an equation like \( \frac{d^2 y}{dx^2} + \mathrm{e}^{-y} = 0 \), we examine the second derivative. For a stationary value where \( \frac{dy}{dx} = 0 \), the given equation modifies to \( \frac{d^2 y}{dx^2} = -\mathrm{e}^{-y} \). Since \( \mathrm{e}^{-y} \) is positive, \( \frac{d^2 y}{dx^2} < 0 \). This indicates the function has a local maximum. Understanding local maxima helps in various optimization problems and in analyzing the behavior of functions.