Question

Consider the following functions and express the relationship between a small change in \(x\) and the corresponding change in \(y\) in the form \(d y=f^{\prime}(x) d x\) $$f(x)=\ln (1-x)$$

Short answer

Question: Given the function \(f(x)=\ln (1-x)\), find the relationship between a small change in \(x\) and the corresponding change in \(y\), and express the result in the form \(dy = f'(x) dx\). Answer: For the function \(f(x)=\ln (1-x)\), the relationship between a small change in \(x\) and the corresponding change in \(y\) is given by the equation \(dy = f'(x) dx = \frac{-1}{1-x} dx\).

Step-by-step solution

Identify the function to differentiate

Begin by identifying the function to differentiate, which is: $$ f(x)=\ln (1-x). $$

Differentiate the function with respect to x

To find the derivative, we will have to use the chain rule since the function is of the form \(\ln(u)\), where \(u=1-x\). The chain rule states that: $$ \frac{d}{dx}[\ln(u)]=\frac{1}{u}\frac{du}{dx}. $$Differentiating \(f(x)\) with respect to \(x\) gives: $$ \begin{aligned} f^{\prime}(x) &= \frac{d}{dx}[\ln(1-x)] \\ &= \frac{1}{1-x}\frac{d}{dx}(1-x). \end{aligned} $$

Differentiate the inside function

Now we need to find the derivative of the inside function, which is \(1-x\): $$ \frac{d}{dx}(1-x)=-1. $$

Complete the chain rule process

With the derivative of the inside function, we can now complete the chain rule process: $$ \begin{aligned} f^{\prime}(x) &= \frac{1}{1-x}\cdot(-1) \\ &= \frac{-1}{1-x}. \end{aligned} $$

Express the relationship between dx and dy

Now that we have the derivative of the function, we can express the relationship between a small change in \(x\) and the corresponding change in \(y\) in the form \(dy = f'(x) dx\): $$ dy = f'(x) dx = \frac{-1}{1-x}dx. $$This equation tells us how a small change in \(x\) affects the value of \(y\), given the function \(f(x)=\ln(1-x)\).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus that deals with how we can find the rate at which one quantity changes with respect to another. When we differentiate a function, we are essentially looking to find the derivative, which is the slope of the tangent line to the function's graph at any given point. This slope tells us how steep the graph is at that point, which in turn informs us about how the function value increases or decreases as we make small changes to the input value.

For example, in physics, differentiation is used to find the rate of change of velocity with respect to time, which is acceleration. In economics, it can describe how the cost might change as production increases. The process of differentiation can apply to a vast array of functions - polynomial, exponential, trigonometric, and more. In each case, rules and formulas have been established to guide how we approach finding the derivative of a function.

Natural Logarithm

The natural logarithm, denoted as \(\ln(x)\), is a special type of logarithm that uses the mathematical constant \(e\) as its base. The number \(e\) is approximately 2.71828 and is the base of the natural logarithms due to its unique properties in calculus, notably in relation to the concept of continuous growth. Logarithms, in general, answer the question: to what power must we raise the base number to obtain another number? For natural logarithms, it translates to 'to what power must we raise \(e\) to get \(x\)?'

The function \(\ln(x)\) is the inverse of the exponential function \(e^x\), and it plays a significant role in many areas of calculus, including integration and solving differential equations. A key property of the natural logarithm is that \(\ln(e) = 1\), and more generally, that \(\ln(e^x) = x\). Its derivative, which we will discuss in the next section, is particularly notable for its simplicity and important applications.

Derivative of Logarithmic Function

The derivative of a logarithmic function, specifically the natural logarithm, has a unique and relatively simple form. If you have \(f(x) = \ln(x)\), the derivative of that function, denoted by \(f'(x)\) or \(\frac{df}{dx}\), is \(\frac{1}{x}\). This elegantly simple derivative indicates that the slope of the tangent line to the graph of \(\ln(x)\) at a point \(x\) is \(\frac{1}{x}\).

When dealing with more complex natural logarithmic functions, such as \(f(x) = \ln(g(x))\), where \(g(x)\) is another function of \(x\), the chain rule must be utilized. The chain rule is a method in calculus for finding the derivative of a composition of functions. It states that the derivative of \(\ln(g(x))\) is \(\frac{1}{g(x)}\cdot g'(x)\), where \(g'(x)\) is the derivative of \(g(x)\). This allows us to handle functions where the natural logarithm is applied to a more complicated argument than just \(x\). The example exercise demonstrates the application of the chain rule to the logarithmic function \(f(x) = \ln(1 - x)\), illustrating how changes in \(x\) result in changes in \(y\), identified through the function's derivative.