Question

Find \(D_{x} y\) using the rules of this section. $$ y=\frac{2 x-1}{x-1} $$

Short answer

The derivative is \(\frac{-1}{(x-1)^2}\).

Step-by-step solution

Identify the Quotient Rule

The function given is a fraction, so the quotient rule of differentiation is applicable. The quotient rule states that if you have a function in the form \( \frac{u}{v} \), then its derivative is given by \( \frac{v \cdot u' - u \cdot v'}{v^2} \). Here, \( u = 2x - 1 \) and \( v = x - 1 \).

Differentiate Numerator and Denominator

Find the derivative of the numerator \( u = 2x - 1 \), which is \( u' = 2 \). Find the derivative of the denominator \( v = x - 1 \), which is \( v' = 1 \).

Apply the Quotient Rule

Substitute \( u, u', v, \) and \( v' \) into the quotient rule formula: \[ D_x y = \frac{(x - 1) \cdot 2 - (2x - 1) \cdot 1}{(x - 1)^2} \] Simplify to get \[ D_x y = \frac{2x - 2 - 2x + 1}{(x - 1)^2} \].

Simplify the Result

Continue simplifying: \[ D_x y = \frac{-1}{(x - 1)^2} \]. This is the derivative of the given function with respect to \( x \).

Key concepts

Differentiation

Differentiation is a fundamental concept in calculus. It refers to the process of finding the derivative of a function, which represents the function's rate of change. In essence, differentiation helps us understand how a function behaves as its input values change. When we differentiate a function related to real-world problems, it allows us to predict or describe its behavior. For example, if you have a function like speed as a function of time, differentiating this function will give you acceleration, which tells you how fast the speed is changing. Differentiation involves applying specific rules, such as the power rule, product rule, and quotient rule, depending on the form of the function.

Calculus

Calculus is a branch of mathematics that primarily deals with change. It has two major branches: differential calculus and integral calculus. In this article, we're focusing on differential calculus, which is concerned with understanding how functions change and finding their derivatives. Calculus enables us to solve complex problems involving dynamic systems across various fields such as physics, engineering, economics, and beyond. It allows us to calculate things like velocity and acceleration, optimize functions in business and physics, and model real-world phenomena like population growth. In our exercise, we're using techniques from differential calculus, specifically the quotient rule, to find how two polynomials, divided by one another, change.

Derivative

The derivative of a function is a measure of how the function value changes as its input changes. Simply put, it tells us the slope of the function at any given point. Derivatives help us find rates of change and solve optimization problems.The notations for derivatives include:

• Greek symbols such as \( \delta \) or \( \Delta \), which indicate a small change or difference.
• The prime notation \( f'(x) \) or \( y' \), which are often used to denote the first derivative of a function \( f \) or \( y \).
• Leibniz notation \( \frac{dy}{dx} \), which expresses the derivative of \( y \) with respect to \( x \).

The exercise involves finding the derivative of a rational function using the quotient rule, which is handy for functions that are ratios of two differentiable functions.

Simplification

Simplification in the context of calculus often involves reducing expressions to their simplest form to make them more understandable or to reveal important features of the function. After applying differentiation rules like the quotient rule, we usually perform simplification to clean up the result.In our exercise, the final step involved simplifying the derived expression:i.e., simplifying from:\[ D_x y = \frac{2x - 2 - 2x + 1}{(x - 1)^2} \]to:\[ D_x y = \frac{-1}{(x - 1)^2} \]This step is crucial since it helps eliminate unnecessary complexity, providing a clearer representation of the function's derivative. Simplification is a necessary skill, helping you check and confirm correctness in your calculus problems.