Question

How do you find the derivative of the quotient of two functions that are differentiable at a point?

Short answer

Question: Apply the Quotient Rule to find the derivative of the function: \[y = \frac{x^2 + 3x - 2}{x - 1}\] Answer: Step 1: Identify the functions u and v In the given function, \[y = \frac{x^2 + 3x - 2}{x - 1}\], we can identify the functions u and v as follows: - u(x) = x^2 + 3x - 2 - v(x) = x - 1 Step 2: Find the derivatives u' and v' Now, we differentiate u(x) and v(x) with respect to x: - u'(x) = 2x + 3 (Differentiating u(x) = x^2 + 3x - 2) - v'(x) = 1 (Differentiating v(x) = x - 1) Step 3: Use the Quotient Rule formula Using the Quotient Rule formula, \[y' = \frac{u'v - uv'}{v^2}\], plug in the values of u, v, u', and v': \[y' = \frac{(2x + 3)(x - 1) - (x^2 + 3x - 2)(1)}{(x - 1)^2}\] Step 4: Simplify the resulting expression Now, simplify the expression for the derivative of y: \[y' = \frac{(2x^2 + x - 3) - (x^2 + 3x - 2)}{(x - 1)^2}\] \[y' = \frac{2x^2 + x - 3 - x^2 - 3x + 2}{(x - 1)^2}\] \[y' = \frac{x^2 - 2x - 1}{(x - 1)^2}\] Thus, the derivative of the given function is: \[y' = \frac{x^2 - 2x - 1}{(x - 1)^2}\].

Step-by-step solution

Identify the functions u and v

Given the quotient of two functions: \[y = \frac{u}{v}\] Identify the functions u and v in the given quotient function.

Find the derivatives u' and v'

Differentiate both u and v with respect to x (or the relevant variable) individually. Call their derivatives u' and v', respectively.

Use the Quotient Rule formula

Following the Quotient Rule formula: \[y' = \frac{u'v - uv'}{v^2}\] Plug in the values of u, v, u', and v' into the formula, and simplify if necessary.

Simplify the resulting expression

Simplify the expression for the derivative to its most simplified form, if at all possible. This may require techniques such as factorization, canceling out common terms, or combining like terms. Following these steps, you will find the derivative of the quotient of two functions that are differentiable at a point, according to the Quotient Rule.

Key concepts

Differentiable Functions

Before we delve into the intricacies of derivatives, it’s pivotal to understand the concept of differentiable functions. A function is said to be differentiable at a point if it has a derivative at that point. This means that the graph of the function has a well-defined tangent at that point, and hence, a specific slope. In a broader sense, if a function is differentiable at every point in an interval, then it has a derivative for every point in that interval, and we can say the function is differentiable on that interval.

It is crucial for students to remember that not all functions are differentiable everywhere. Classic examples include functions that have sharp turns or cusps, like the absolute value function at zero, or vertical tangents, such as the function defined by \( y = x^{1/3} \) at zero. Before applying the Quotient Rule, it’s essential to ensure both the numerator and denominator functions are differentiable, particularly at the points of interest.

Understanding differentiability is paramount as it ensures the reliability of derivatives and the consequent analysis of the function’s behavior.

Finding Derivatives

Finding derivatives is a fundamental process in calculus that involves calculating the rate at which one quantity changes with respect to another. It can be seen as an application of limits and closely resembles the slope of the tangent line to a function's graph at a specific point. The derivative, often denoted as \( f'(x) \), is a central tool that enables us to understand how functions behave, predict future trends, and solve an array of problems in various disciplines such as physics, economics, and engineering.

To find derivatives, multiple rules and formulas have been developed to make the process more systematic. Among them, the Quotient Rule is particularly useful when dealing with quotients of functions, as mentioned in the exercise above. The beauty of the Quotient Rule is that it provides a direct way to find the derivative of a division of two differentiable functions without the need to rewrite or manipulate the function into a different form.

When memorizing the Quotient Rule, which states that \( (\frac{u}{v})' = \frac{u'v - uv'}{v^2} \), remember that \( u \) and \( v \) stand for differentiable functions, and \( u' \) and \( v' \) are their derivatives, respectively. It's a matter of plugging in the right values and simplifying.

Pro Tip:

Always verify that \( v \) is never zero, as division by zero is undefined, which would mean that the derivative doesn't exist at those points.

Simplifying Expressions

Once we have used the Quotient Rule to find the derivative, we are often left with an expression that is not in its simplest form. Simplifying expressions is a skill that enhances clarity and can make it easier to identify critical features of the derivative, such as its zeros and signs. This, in turn, helps in understanding the behavior of the original function.

Simplification can involve a range of techniques: canceling common factors, expanding products, and combining like terms, to name a few. It's about expressing the derivative in a way that is easier to interpret and use, be it for sketching a graph or calculating values.

Common Mistakes:

One familiar stumbling block is canceling terms incorrectly. Be vigilant to only cancel factors that are common to both the numerator and denominator, and remember that terms added or subtracted in the numerator cannot be canceled with those in the denominator.

Another prudent practice is to check for opportunities to factor the expression. Factoring can often reveal zeros of the derivative that are critical in the study of the function’s extrema. Keeping the derivative expression clean and minimal is not just an aesthetic choice, but a practical one that makes subsequent calculus operations much easier to perform.