Question

Derivatives of products and quotients Find the derivative of the following functions by first expanding or simplifying the expression. Simplify your answers. $$y=\frac{x^{2}-2 a x+a^{2}}{x-a} ; a \text { is a constant. }$$

Short answer

Answer: The derivative of the given function is 1.

Step-by-step solution

Simplify function expression

We can simplify the given function as follows: \(y=\frac{x^{2}-2 a x+a^{2}}{x-a}\) By factoring the numerator, $$= \frac{(x-a)^{2}}{x-a}$$ After canceling out \((x-a)\) in both numerator and denominator, $$y = x-a$$ So the simplified function is \(y = x-a\).

Differentiate the simplified function

Now we will differentiate the simplified function \(y = x-a\) with respect to \(x\). $$\frac{dy}{dx} = \frac{d}{dx}(x-a)$$ As derivatives are linear operators, $$= \frac{d}{dx}x - \frac{d}{dx}a$$

Apply derivative rules

Now we need to apply the basic differentiation rules to find the derivatives: The derivative of a constant with respect to its variable is zero: $$\frac{d}{dx}a = 0$$ and The derivative of x with respect to itself is 1: $$\frac{d}{dx}x = 1$$ Therefore, $$\frac{dy}{dx} = 1 - 0$$

Simplify the result

Finally, we can simplify the result and get the final derivative of the function: $$\frac{dy}{dx} = 1$$ So, the derivative of the given function is 1.

Key concepts

Simplifying Expressions

When tackling complex mathematical problems, simplifying expressions plays a crucial role in making the solutions more approachable and manageable. Simplification often involves reducing to lower terms by canceling out factors, combining like terms, and factoring polynomials.

Consider the provided exercise where we have a fraction with a quadratic expression in the numerator and a linear term in the denominator. By observing the quadratic as a perfect square, it can be factored into \( (x-a)^2 \) and then greatly simplified by canceling out the common term \( (x-a) \) in the numerator and denominator. This leaves us with \( y = x - a \) – a much simpler and more pliable expression for further operations such as differentiation.

Differentiation Rules

Understanding differentiation rules is essential for calculus students. Differentiation rules allow us to determine the rate at which a quantity changes. In problems involving derivatives of products and quotients, knowing these rules can simplify the process considerably.

For the simplified expression \( y = x - a \) obtained from the previous simplification, we applied basic differentiation rules: the derivative of a constant (\(a\)) with respect to \(x\) is zero, and the derivative of \(x\) with respect to itself is one. These rules streamlined our differentiation process, leading to a straightforward derivative, \( \frac{dy}{dx} = 1 \).

Linear Operators in Calculus

In calculus, linear operators are tools that let us break down complex operations into simpler parts that adhere to the principles of linearity – additivity and homogeneity. Differentiation is one such linear operator. It allows us to separate expressions into individual terms and treat each term independently.

For example, when differentiating \( y = x - a \), we can split the differentiation process into \( \frac{d}{dx}x \) and \( -\frac{d}{dx}a \). By treating these separately, we simplify the process and ensure that our operations on each term conform to differentiation rules. This approach underlines the power and elegance of linear operators in solving calculus problems.

Factoring Polynomials

Factoring polynomials is a technique used to simplify expressions where a polynomial is expressed as the product of its factors. This can significantly ease subsequent algebraic manipulation, including differentiation or integration. In the initial exercise, we saw a quadratic polynomial \( x^2 - 2ax + a^2 \) factored into \( (x - a)^2 \) due to the perfect square relationship.

Being adept at recognizing common polynomial patterns, such as perfect squares, difference of squares, or sum and difference of cubes, can be of immense help. These patterns can turn daunting, lengthy problems into manageable ones through simplification, which is especially beneficial prior to performing operations like differentiation.