Problem 85 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) is differentiable. Find an expression for the derivative of \(y\) at \(x=2\), assuming that \(f(2)=\) \(-1\) and \(f^{\prime}(2)=1\). $$ =\frac{x^{2}+4 f(x)}{f(x)} $$

Step-by-Step Solution

Answer

The derivative of the function at \( x = 2 \) is \(-8\).

Step 1: Identify the function

The given function is \( y = \frac{x^2 + 4f(x)}{f(x)} \). Your task is to find the derivative of this function at \( x = 2 \).

Step 2: Apply the quotient rule

To differentiate \( y = \frac{u}{v} \), where \( u = x^2 + 4f(x) \) and \( v = f(x) \), apply the quotient rule: \( y' = \frac{v \cdot u' - u \cdot v'}{v^2} \).

Step 3: Differentiate the numerator

Find \( u' \), the derivative of \( x^2 + 4f(x) \). Use the sum and chain rule: \( u' = 2x + 4f'(x) \).

Step 4: Differentiate the denominator

Differentiate \( v = f(x) \): \( v' = f'(x) \).

Step 5: Plug in x = 2

Substitute \( x = 2 \), \( f(2) = -1 \), and \( f'(2) = 1 \) into the derivatives: - \( u' = 2(2) + 4(1) = 4 + 4 = 8 \) - \( v' = 1 \)

Step 6: Evaluate the derivative using the quotient rule

Substitute the values into the quotient rule: \( \frac{(-1)(8) - ((2^2 + 4(-1))(1))}{(-1)^2} \). Simplify: \[ y' = \frac{-8 - (4 - 4)}{1} = -8 \].

Key Concepts

Quotient Rule

The quotient rule is essential when finding the derivative of a function represented as a division of two differentiable functions. If you have two functions, say \( u(x) \) and \( v(x) \), and you want to find the derivative of their quotient \( \frac{u(x)}{v(x)} \), the quotient rule provides a systematic way to do it.
The formula is:

• \( y' = \frac{v(x) \, u'(x) - u(x) \, v'(x)}{(v(x))^2} \)

This ensures that you get the correct derivative by carefully weighing the interaction of \( u \) and \( v \).
When applying it to our problem:

• \( u(x) = x^2 + 4f(x) \)
• \( u'(x) = 2x + 4f'(x) \)
• \( v(x) = f(x) \)
• \( v'(x) = f'(x) \)

Substituting back into the quotient rule helps us compute the derivative step by step.

Chain Rule

The chain rule is a powerful technique in calculus used for differentiating compositions of functions. When you have a function of a function, such as \( f(g(x)) \), the chain rule helps in finding its derivative efficiently.
If \( y = f(g(x)) \), the derivative with respect to \( x \) is given by:

• \( y' = f'(g(x)) \, g'(x) \)

This highlights how changes in the inner function \( g(x) \) affect the outer function \( f \).
In our exercise, the chain rule is applied within the expression \( 4f(x) \) found in the numerator \( x^2 + 4f(x) \).
We differentiate the inner function \( f(x) \) using \( f'(x) \) and then multiply it by the constant multiplier 4.
This gives us \( 4f'(x) \), capturing how \( f(x) \) responds to changes in \( x \).

Derivatives

Derivatives represent how a function changes as its input changes. They are foundational in calculus and provide insightful information about the behavior of functions.
A derivative is often denoted as \( f'(x) \) or \( \frac{dy}{dx} \), which tells you the slope or rate of change of a function at a particular point.
In practice, finding derivatives involves using rules like the quotient rule, product rule, power rule, and chain rule, each suited for different types of functions.
In this exercise:

• We needed to differentiate \( y = \frac{x^2 + 4f(x)}{f(x)} \) at \( x = 2 \)
• The context provided \( f(2) = -1 \) and \( f'(2) = 1 \), which were crucial for substitution after applying differentiation rules.
• The final derivative value of \(-8\) conveys the rate at which \( y \) is changing at that specific point on its curve.

Understanding derivatives helps with problems ranging from physics to economics, as it deals with dynamic changes rather than static values.