Problem 89 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable at x. Find an expression for the derivative of y in terms of \(f(x), g(x), f^{\prime}(x)\), and \(g^{\prime}(x) .\) \(y=\frac{f(x)}{[g(x)]^{2}}\)

Step-by-Step Solution

Answer

The derivative is given by \( y' = \frac{f'(x)[g(x)]^2 - 2f(x)g(x)g'(x)}{[g(x)]^4} \).

Step 1: Recognize the Problem Type

The function given is a quotient, so we need to use the quotient rule for differentiation.

Step 2: Quotient Rule Application

The quotient rule states that if \( y = \frac{u}{v} \), then the derivative \( y' = \frac{u'v - uv'}{v^2} \). Here \(u = f(x)\) and \(v = [g(x)]^2\).

Step 3: Calculate Derivatives of Numerator and Denominator

We need the derivatives of the numerator and the denominator. Since \(u = f(x)\), \(u' = f'(x)\). For \(v = [g(x)]^2\), use the chain rule to find \(v'\): \(v' = 2g(x)g'(x)\).

Step 4: Apply the Quotient Rule

Substitute the expressions into the quotient rule formula: \[ y' = \frac{f'(x) \cdot [g(x)]^2 - f(x) \cdot 2g(x)g'(x)}{([g(x)]^2)^2} \]

Step 5: Simplify the Expression

Simplifying the expression, we have: \[ y' = \frac{f'(x)[g(x)]^2 - 2f(x)g(x)g'(x)}{[g(x)]^4} \].

Key Concepts

Understanding the Quotient Rule

The quotient rule is a fundamental tool in calculus, specifically in differentiation, used when dealing with ratios of two differentiable functions. If you have a function that can be expressed as the division of one function by another, this is your go-to method. The quotient rule states:\[ y = \frac{u}{v} \quad \Rightarrow \quad y' = \frac{u'v - uv'}{v^2} \]Here, else \(u\) and \(v\) represent differentiable functions of \(x\). The prime symbols \(u'\) and \(v'\) indicate their respective derivatives. This method ensures that the differentiation process is both accurate and efficient when applied to fractions of functions. In the example from the exercise, the function \(y = \frac{f(x)}{[g(x)]^2}\) is structured perfectly for the quotient rule, making it necessary to apply it to find the derivative correctly.

Decoding the Chain Rule

The chain rule is an essential calculus tool when differentiating composite functions. A composite function is when one function is nested within another, resembling layers of an onion. The chain rule helps in peeling away these layers correctly to find the derivative. The chain rule states that if you have a composite function \(z(x) = h(g(x))\) then:\[ z'(x) = h'(g(x)) \cdot g'(x) \]This means you first differentiate the outer function (\(h\)) with respect to its inner function (\(g(x)\)), and then multiply by the derivative of the inner function.
The example in the exercise involves \(v = [g(x)]^2\), which can be viewed as a composite function: the square of \(g(x)\). Applying the chain rule here involves taking the derivative of the squaring function and multiplying it by the derivative of \(g(x)\).
Thus, \(v' = 2g(x) \, g'(x)\). This differentiation method simplifies breaking down complicated functions into manageable pieces.

Mastering Derivatives of Functions

Derivatives are the cornerstone of calculus, representing the rate of change of a function. They provide critical insights into the function's behavior, such as velocities, slopes, and trends. The derivative of a function \(f(x)\) is commonly denoted as \(f'(x)\). There are standard rules and techniques to find the derivative of various forms of functions:

• **Basic Functions:** The derivative of \(x^n\) is \(nx^{n-1}\).
• **Exponential Functions:** For \(e^x\), the derivative is \(e^x\).
• **Trigonometric Functions:** The derivative of \(\sin(x)\) is \(\cos(x)\), and the derivative of \(\cos(x)\) is \(-\sin(x)\).

In the context of the exercise, differentiating \(f(x)\) and \(g(x)\) involves using these principles to determine \(f'(x)\) and \(g'(x)\). By mastering these derivative rules, you can simplify complex expressions and gain profound insights into the dynamics of mathematical functions.