91Ó°ÊÓ

The derivative is a fundamental concept in calculus, representing the rate at which a function changes as its input changes. For a function like \( f(x) = \frac{1}{(1+x)^2} \), the derivative helps us understand how \( f(x) \) behaves for small changes around a given point, \( a \). For linear approximation, the derivative at point \( a \), noted as \( f'(a) \), indicates the slope of the tangent line to \( f(x) \) at that point.
The process of finding the derivative of \( \frac{1}{(1+x)^2} \) involves the differentiation of a power law function. For this, we use the rule that describes how to differentiate \( (g(x))^n \), specifically leading to \(-2(1+x)^{-3}\) when applying the chain rule. This result illustrates that for every slight increase in \( x \), \( f(x) \) decreases at a certain rate.
Understanding derivatives is crucial because they give us the gradient of a function, enabling us to make predictions about the function's behavior near specific points. These predictions are the essence of linear approximations.

The chain rule is a powerful technique for differentiating complex functions that are composed of simpler functions. It allows us to find the derivative of a function that is a function of another function. For the function \( f(x) = (1+x)^{-2} \), the chain rule simplifies the process.
In this exercise, the function \( (1+x)^{-2} \) is a composition of an outer function, \( u^n \), and an inner function, \( u = 1+x \). The chain rule requires taking the derivative of the outer function with respect to the inner function and then multiplying it by the derivative of the inner function itself. * The derivative of \( u^n \) where \( n = -2 \) is \( -2u^{-3} \).
* The derivative of \( u = 1+x \) is \( 1 \), since the derivative of \( x \) is \( 1 \) and the derivative of a constant is \( 0 \).
Combining these uses of the chain rule gives us \( f'(x) = -2(1+x)^{-3} \). This rule is indispensable in calculus, especially when dealing with functions that comprise several layers.

Function evaluation involves calculating the value of a function for specific inputs. It is a straightforward yet essential procedure in calculus, particularly when applying concepts such as linear approximation.
To evaluate a function at a point, simply substitute the specific value into the function formula. In our exercise, we needed to evaluate \( f(x) = \frac{1}{(1+x)^2} \) at \( a = 0 \). Substituting gives us:* \( f(0) = \frac{1}{(1+0)^2} = 1 \)
Function evaluation also requires calculating the derivative at this point. In this case, substituting \( a = 0 \) into the derivative \( f'(x) = -2(1+x)^{-3} \) yields:* \( f'(0) = -2(1+0)^{-3} = -2 \)
Both function evaluation and derivative evaluation are key to determining linear approximations. They help in simplifying computations and making precise predictions about function behavior for values near \( a = 0 \).