Problem 39 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable. Find \(\frac{d}{d x} \frac{[f(x)]^{2}}{g(2 x)+2 x}\)

Step-by-Step Solution

Answer

The derivative is \( \frac{2(g(2x)+2x)f(x)f'(x) - [f(x)]^2(2g'(2x) + 2)}{(g(2x) + 2x)^2} \).

Step 1: Identify the Type of Function

The function in question is a quotient: \( \frac{[f(x)]^2}{g(2x)+2x} \). To differentiate this, we use the quotient rule.

Step 2: Apply the Quotient Rule

The quotient rule states that if you have a function \( \frac{u(x)}{v(x)} \), its derivative is given by \( \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \). Here, \( u(x) = [f(x)]^2 \) and \( v(x) = g(2x) + 2x \).

Step 3: Differentiate the Numerator Using the Chain Rule

The numerator \( u(x) = [f(x)]^2 \). By the chain rule, its derivative is \( u'(x) = 2f(x)f'(x) \).

Step 4: Differentiate the Denominator Using the Chain Rule

The denominator \( v(x) = g(2x) + 2x \). For \( g(2x) \), apply the chain rule: its derivative is \( 2g'(2x) \). Thus, \( v'(x) = 2g'(2x) + 2 \).

Step 5: Substitute into the Quotient Rule Formula

Substitute the derivatives into the quotient rule formula: \[ \frac{d}{dx} \left( \frac{[f(x)]^2}{g(2x)+2x} \right) = \frac{(g(2x) + 2x)(2f(x)f'(x)) - ([f(x)]^2)(2g'(2x) + 2)}{(g(2x) + 2x)^2} \].

Step 6: Simplify the Expression

Simplify the expression: \[ \frac{2(g(2x)+2x)f(x)f'(x) - [f(x)]^2(2g'(2x) + 2)}{(g(2x) + 2x)^2} \].

Key Concepts

Quotient Rule in Differentiation

When working with derivatives, the quotient rule is essential anytime you have a function that involves one function divided by another, i.e., a fraction or a "quotient."
This rule helps you find the derivative of such fractions precisely and efficiently.

• The general form of a function requiring the quotient rule is \(\frac{u(x)}{v(x)}\), where both \(u(x)\) and \(v(x)\) are differentiable functions.
• The quotient rule formula is: \[ \left( \frac{u(x)}{v(x)} \right)' = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} \]
• This rule stems from the product and chain rules and allows calculating derivatives, especially when both parts of the fraction are more complex functions.

You begin by identifying which part of your function is \(u(x)\) and which is \(v(x)\).
In this exercise, \(u(x) = [f(x)]^2\) and \(v(x) = g(2x) + 2x\).
Once identified, finding their derivatives allows for direct application of the quotient rule formula.

Understanding the Chain Rule

The chain rule is another crucial tool in calculus differentiation, especially for composite functions, where one function is nested inside another.
This rule allows us to differentiate layered functions piece by piece, ensuring accuracy and simplicity.

• A composite function has the form \(f(g(x))\) or similar, where \(f(x)\) is a function applied to \(g(x)\).
• The chain rule formula is: \[ \frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x) \]
• This helps when differentiating through multiple function "layers," such as polynomials and trigonometric functions combined.

In this problem, the chain rule is applied to both the numerator\([f(x)]^2\) and the denominator \(g(2x)\).
For \([f(x)]^2\), recognizing it as \((f(x))^2\) makes it clear how to apply the chain rule, yielding \(2f(x)f'(x)\).
For \(g(2x)\), the chain involves \(2x\) as the inner function, resulting in \(2g'(2x)\).

Basic Derivatives of Functions

Understanding how to find the derivatives of primary functions is foundational in calculus.
The goal is to know the basic steps and formulas to apply when face-to-face with any function.

• Common functions include powers like \(x^n\), where the derivative formula is: \[ \frac{d}{dx}(x^n) = nx^{n-1} \]
• The exponential function \(e^x\) or \(a^x\) and corresponding derivatives are straightforward, like: \(e^x\) remains \(e^x\), and \(a^x\) becomes \(a^x \ln(a)\).
• Derivatives of trigonometric functions such as \(\sin(x)\) and \(\cos(x)\) also follow specific patterns: \(\frac{d}{dx}\sin(x) = \cos(x)\), and \(\frac{d}{dx}\cos(x) = -\sin(x)\).

Combining these basics with rules like the product, quotient, and chain rules helps handle more complex problems.
As in our exercise, knowing \(f(x)^2\) directly shows the need for both basic derivative knowledge and further application of the chain rule.