Problem 40 - Calculus for Biology and Medicine

Question

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

Step-by-Step Solution

Answer

The derivative of \( y \) at \( x = 1 \) is \( \frac{1}{2} \).

Step 1: Apply the Product Rule

The function is given as \( y = \frac{x f(x)}{2} \), which can be rewritten as \( y = \frac{1}{2} x f(x) \). To find the derivative \( y' \), we first need to apply the product rule to \( x f(x) \). According to the product rule, \( (uv)' = u'v + uv' \). Here, let \( u = x \) and \( v = f(x) \). So, \( u' = 1 \) and \( v' = f'(x) \).

Step 2: Differentiate Using the Product Rule

Using the product rule: \((x f(x))' = x' f(x) + x f'(x) = f(x) + x f'(x)\). Then multiply the entire expression by \(\frac{1}{2}\), since \( y = \frac{1}{2} x f(x) \), to get \( y' = \frac{1}{2} (f(x) + x f'(x)) \).

Step 3: Substitute Given Values

Now, substitute the given values \( f(1) = 2 \) and \( f'(1) = -1 \) into the derivative expression \( y'(x) = \frac{1}{2} (f(x) + x f'(x)) \). This gives us \( y'(1) = \frac{1}{2} (f(1) + 1 * f'(1)) = \frac{1}{2} (2 + 1*(-1)) \).

Step 4: Simplify the Expression

Simplify the expression to find the numerical value for \( y'(1) \): \( y'(1) = \frac{1}{2} (2 - 1) = \frac{1}{2}(1) = \frac{1}{2} \).

Key Concepts

Product Rule

The product rule is a fundamental principle in calculus used to differentiate functions that are products of two or more separate functions. When you have a function defined by two functions multiplied together, such as \( y = u(x) \cdot v(x) \), the product rule provides a method to find its derivative. For this case, the product rule is expressed as:

• \( (uv)' = u'v + uv' \)

This formula tells us that the derivative of a product of two functions \( u \) and \( v \) equals the derivative of \( u \) times \( v \), plus \( u \) times the derivative of \( v \). By using the product rule, you accurately "distribute" the differentiation process over each term in the product.

In the problem we are tackling, \( y = \frac{1}{2}x f(x) \), we express it as a product \( u = x \) and \( v = f(x) \). Therefore, \( u' = 1 \) and \( v' = f'(x) \), allowing the use of the product rule to simplify the derivative process.

Derivative Evaluation

Once the derivative is formulated using the product rule, it becomes critical to evaluate it at specific points to find the slope of the tangent line at those points. This step transforms the abstract derivative into concrete information about the behavior of the function at a given point. Evaluating derivatives requires substituting the specific values for the variables and the derivatives given in the problem.

In our scenario, we had to find the derivative at \( x = 1 \). Given \( f(1) = 2 \) and \( f'(1) = -1 \), we substituted these directly into our derivative formula:

• \( y'(x) = \frac{1}{2} (f(x) + x f'(x)) \)
• \( y'(1) = \frac{1}{2} (2 + 1 \cdot (-1)) \)

By substituting and simplifying, you transform theoretical knowledge into practical values, obtaining \( y'(1) = \frac{1}{2}\). This value represents the slope of the tangent to the curve \( y \) at \( x = 1 \).

Function Differentiability

Differentiability is an essential concept in calculus that describes a function's ability to be differentiated. A function is said to be differentiable at a point if it has a derivative at that point, meaning it can be represented by a tangent line that smoothly touches the curve. More formally, a function \( f(x) \) is differentiable at \( x = a \) if the following limit exists:

• \( \lim_{h \to 0} \frac{f(a + h) - f(a)}{h} \)

Differentiability requires the function to be continuous, meaning there should be no abrupt changes, holes, or jumps at the point of interest.

In our specific problem, we operate under the assumption that \( f(x) \) is differentiable. This guarantees that all of the derivative manipulations, such as applying the product rule and evaluating the derivative, can validly occur. Differentiability at \( x = 1 \) ensures that not only does \( f(x) \) have a well-defined tangent line at this point, but it also supports calculating \( f'(1) \) and provides a necessary foundation for analyzing other derivative-related properties of the function.