Problem 37 - 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 \(f^{\prime}(1)=-1\) \(y=2 x f(x)\)

Step-by-Step Solution

Answer

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

Step 1: Apply the Product Rule

The function given is the product of two functions: \(2x\) and \(f(x)\). To find the derivative \(y'\), we will apply the product rule. The product rule states that \((uv)' = u'v + uv'\), where \(u = 2x\) and \(v = f(x)\). Thus, \(u' = 2\) and \(v' = f'(x)\).

Step 2: Substitute into the Product Rule

Using the product rule, compute the derivative: \[ y' = (2x) imes f'(x) + 2 imes f(x) \]

Step 3: Substitute Given Values

Now, substitute the given values at \(x = 1\): - \(f(1) = 2\) - \(f'(1) = -1\)Compute \[ y' = 2 imes 1 imes (-1) + 2 imes 2 = -2 + 4\]

Step 4: Calculate the Result

After substituting the values and calculating, we find \[ y' = 2 \] at \(x = 1\).

Key Concepts

Understanding the Product Rule

When we have two functions being multiplied together, like in the function \(y = 2x f(x)\), we need the product rule to differentiate them. The product rule is a handy tool in calculus for finding derivatives of products of two functions. Here's the key formula for the product rule:

• If \(u(x)\) and \(v(x)\) are functions of \(x\), then the derivative of their product \(uv\) is \((uv)' = u'v + uv'\).
• In our example, we set \(u = 2x\) and \(v = f(x)\).
• This means the derivative \(u' = 2\) because the derivative of \(2x\) with respect to \(x\) is just the coefficient, and \(v' = f'(x)\), which is given.

Applying the product rule helps us manage problems where two functions are multiplied together, simplifying complex calculations into more manageable pieces. That's the beauty of the product rule!

Unraveling the Derivative

A derivative measures how a function changes as its input changes. It's a core concept in calculus that tells us the rate of change or the slope of the function’s graph at any point. In the context of \(y = 2x f(x)\), we're interested in how \(y\) changes at a particular point, specifically at \(x=1\).
To find the derivative of the product \(2x \times f(x)\), we used the product rule. We found:

• \( (2x f(x))' = 2 f(x) + 2x f'(x) \)

A derivative gives us insight into how quickly the output of a function is changing relative to changes in the input. When calculating derivatives, using rules like the product rule is crucial to simplifying the process. Differentiating can be challenging without these fundamental guidelines.
Once calculated, the derivative can answer questions about the function's behavior at specific points, such as how fast it increases or decreases.

Defining Functions and Their Importance

A function relates an input to an output, following a specific rule. In mathematics, functions are essential because they model relationships between varying quantities.
In our problem, \(f(x)\) represents an unknown function that is differentiable, meaning it has a derivative. Functions can be simple like linear functions (straight lines) or more complex, possibly involving exponentials, logs, or trigonometric operations.

• For our function \(y = 2x f(x)\), we worked with a product of two simpler functions: the linear function \(2x\) and an unknown \(f(x)\).
• The derivative \(f'(x)\) tells us how \(f(x)\) changes with \(x\).

Understanding functions is a stepping stone to many areas in mathematics and science. They provide a way to capture and describe changes and are indispensable in solving real-world problems, making them a fundamental concept in calculus and beyond.