Problem 86 - 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 \(f^{\prime}(2)=1\) \(y=[f(x)]^{2}-\frac{x}{f(x)}\)

Step-by-Step Solution

Answer

The expression for the derivative of \(y\) at \(x=2\) is 1.

Step 1: Write down the expression for the function

The function given is \(y = [f(x)]^2 - \frac{x}{f(x)}\). We will find its derivative using the chain rule and quotient rule.

Step 2: Differentiate the first term

The first term is \([f(x)]^2\). To differentiate this, we use the chain rule: \(\frac{d}{dx}([f(x)]^2) = 2f(x) \cdot f'(x)\).

Step 3: Differentiate the second term using the quotient rule

The second term is \(-\frac{x}{f(x)}\). Using the quotient rule, \(\frac{d}{dx}\left(-\frac{x}{f(x)}\right)\) is \(-\left(\frac{f(x) - x f'(x)}{[f(x)]^2}\right)\).

Step 4: Combine the derivatives

Add the derivatives of both terms together to get the derivative of \(y\). Hence, \(y' = 2f(x)f'(x) - \left(\frac{f(x) - x f'(x)}{[f(x)]^2}\right)\).

Step 5: Evaluate the derivative at \(x=2\) using given values

Substitute \(f(2) = -1\) and \(f'(2) = 1\) into the expression for \(y'\):\[y' = 2(-1)(1) - \left(\frac{-1 - 2 \cdot 1}{[-1]^2}\right) = -2 + \left(\frac{-1 -2}{1}\right) = -2 - (-3)\]\[-2 - (-3) = -2 + 3 = 1\]

Key Concepts

Understanding the Chain Rule

The Chain Rule is a fundamental technique used in calculus for finding the derivative of composite functions. A composite function is formed when one function is nested inside another, such as - \( g(x)=f(h(x)) \).
To differentiate such a function, we use the chain rule: - \( g'(x) = f'(h(x)) \cdot h'(x) \).

This means we first differentiate the outer function, leaving the inner function unchanged, and then multiply by the derivative of the inner function.
In our exercise, the first part \(([f(x)]^2)\) uses the chain rule.

- The outer function is \(u^2\), with derivative \(2u\). - The inner function is \(u = f(x)\).
Thus, when applying the chain rule, the derivative is \(2f(x)\cdot f'(x)\).
The chain rule is invaluable when dealing with layers of functions, simplifying the complex problem of differentiation, step by step.

Mastering the Quotient Rule

The Quotient Rule is specifically meant for differentiating functions that are ratios of two other functions.
If you have a function of the form: - \( h(x) = \frac{a(x)}{b(x)} \), then the derivative of \(h(x)\), denoted as \(h'(x)\), can be found using the formula: - \( h'(x) = \frac{a'(x)b(x) - a(x)b'(x)}{[b(x)]^2} \).

For the expression we are working with, which is the derivative of \(-\frac{x}{f(x)}\), our functions are:- The numerator \(a(x) = -x\) and its derivative \(a'(x) = -1\)- The denominator \(b(x) = f(x)\) with its derivative \(b'(x) = f'(x)\).
By applying the quotient rule, it follows:- The derivative becomes \(-\left(\frac{f(x) - x f'(x)}{[f(x)]^2}\right)\).
The quotient rule saves us time and avoids potential errors when dealing with complex fractions, which could otherwise be cumbersome to differentiate directly.

The Process of Derivative Evaluation

Derivative evaluation involves computing the derivative at a specific point after the general formula of the derivative has been determined.
This is particularly important when solving real-world problems where exact values at specific points are necessary.
When evaluating the derivative of our function \(y\) at \(x=2\), we substitute the given values of the function and its derivative at that point.
We have the function values:

• \(f(2) = -1\)
• \(f'(2) = 1\)

By plugging these into our derived expression \(y' = 2f(x)f'(x) - \left(\frac{f(x) - x f'(x)}{[f(x)]^2}\right)\) results in:

• \(y' = 2(-1)(1) - \left(\frac{-1 - 2 \cdot 1}{[-1]^2}\right)\)
• Which simplifies to \(-2 + 3 = 1\)

This process completes our derivative evaluation, giving us a straightforward numerical result, valuable for interpreting or further application in specific problems.