Problem 91 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable at x. Find an expression for the derivative of y in terms of \(f(x), g(x), f^{\prime}(x)\), and \(g^{\prime}(x) .\) \(y=\sqrt{x} f(x) g(x)\)

Step-by-Step Solution

Answer

The derivative expression is \( y' = \frac{1}{2} x^{-1/2} f(x) g(x) + x^{1/2} (f'(x) g(x) + f(x) g'(x)) \).

Step 1: Identify the derivative rules.

Since the function is a product of multiple functions, we will need to use both the product and chain rules to find the derivative of \( y \). The product rule states that if you have a function \( u(x) v(x) \), its derivative is \( u'(x)v(x) + u(x)v'(x) \). The chain rule will be used for the square root part.

Step 2: Apply the chain rule.

Recognize that \( y = x^{1/2} f(x) g(x) \). Here, the outer function is the square root and the inner function is \( x \). The derivative of the square root is \( \frac{1}{2} x^{-1/2} \).

Step 3: Differentiate using the product rule.

Consider \( u(x) = x^{1/2} \) and \( v(x) = f(x)g(x) \). Use the product rule: \( (uv)' = u'v + uv' \).

Step 4: Differentiate \( u(x) \) and \( v(x) \) separately.

For \( u(x) = x^{1/2} \), we have \( u'(x) = \frac{1}{2}x^{-1/2} \). For \( v(x) = f(x) g(x) \), use the product rule again: \( v'(x) = f'(x)g(x) + f(x)g'(x) \).

Step 5: Combine the derivatives from Steps 3 and 4.

Substitute the derivatives from Steps 4 into the product rule formula: \( y' = \frac{1}{2}x^{-1/2} f(x)g(x) + x^{1/2} (f'(x)g(x) + f(x)g'(x)) \).

Step 6: Simplify the expression.

The expression simplifies to \[ y' = \frac{1}{2} x^{-1/2} f(x) g(x) + x^{1/2} f'(x) g(x) + x^{1/2} f(x) g'(x) \].

Key Concepts

Product Rule

When you encounter a product of two or more functions, the product rule is an essential differentiation technique. It allows us to find the derivative of a product involving differentiable functions. The product rule states:
If you have two functions, say \( u(x) \) and \( v(x) \), the derivative of their product \( y = u(x)v(x) \) is given by the formula: \[(uv)' = u'v + uv'\]

• \( u'(x) \) is the derivative of \( u(x) \)
• \( v'(x) \) is the derivative of \( v(x) \)

These derivatives are then used to form the sum of the two terms. The concept is particularly useful when dealing with functions like \( f(x)g(x) \), ensuring that each part's variation is considered.
For our exercise, the product rule is applied separately after the square root function is addressed.

Chain Rule

The chain rule is pivotal when a function is nested within another function, meaning you have a composition of functions. Primarily used when differentiating expressions like \( f(g(x)) \), the chain rule allows you to "chain" together the derivatives of these functions. Essentially: \[\frac{d}{dx} [f(g(x))] = f'(g(x))g'(x)\] When applying this rule, consider:

• Identify the outer function. In our example, it is the square root: \( x^{1/2} \).
• Recognize the inner function, typically the expression inside, which for us is \( x \).

Apply the chain rule to differentiate the outer function, multiplying it by the derivative of the inner function.
The chain rule simplifies working with composite functions, making the seemingly complex differentiation manageable and logical.

Differentiation Techniques

Differentiation, core to calculus, involves finding the derivative of a function, indicating the rate of change or the slope of the function at any point. Various techniques are available to tackle different forms of functions. Here's a recap:

• Basic Differentiation: Directly finding the derivative using simple algebraic rules.
• Product Rule: As explained, this is used for products of functions.
• Chain Rule: Crucial for composite functions.
• Quotient Rule: Used for functions that are quotients.

In calculus problems, integrating these methods deftly is vital. Each function type in a problem might require a different strategy, like in our exercise involving a product of the square root and two separate functions. Recognizing which rule applies provides clarity and direction.

Calculus Application

Calculus, the branch of mathematics dealing with change, finds applications from school homework to complex physics models. Differentiation sits at its heart, offering tools like the product and chain rules that break down tough problems into solvable parts.
Why is this important? The derivative tells us how functions behave, predicting trends and patterns.
Imagine you're tasked with understanding the growth of a biological population over time. The function describing this growth might be complex, a combination of various factors. Calculus helps unravel these complexities, offering insight into each contributing component's role.
In our specific problem, although simplified, you applied calculus to differentiate \( y = \sqrt{x} f(x) g(x) \), visualizing how each variable impacts the whole. This not only solves math problems, but also trains your mind for broader analytical thinking.