Problem 43 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable at \(x .\) Find an expression for the derivative of \(y .\) \(=[f(x)+2 g(x)] g(x)\)

Step-by-Step Solution

Answer

The derivative of \( y \) is \( f'(x)g(x) + f(x)g'(x) + 4g(x)g'(x) \).

Step 1: Identify the Given Function

The function given is a product of functions: \[ y = [f(x) + 2g(x)]g(x) \].Here, \( u(x) = f(x) + 2g(x) \) and \( v(x) = g(x) \). We need to find the derivative \( \frac{dy}{dx} \).

Step 2: Apply the Product Rule

To differentiate a product of two functions, \( y = u(x) \cdot v(x) \), we use the product rule: \[ \frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x) \].Here, we need to find \( u'(x) \) and \( v'(x) \).

Step 3: Differentiate Each Function

Differentiate \( u(x) = f(x) + 2g(x) \):\[ u'(x) = f'(x) + 2g'(x) \].Differentiate \( v(x) = g(x) \):\[ v'(x) = g'(x) \].

Step 4: Substitute into the Product Rule

Substitute the derivatives found into the product rule formula:\[ \frac{dy}{dx} = (f'(x) + 2g'(x))g(x) + (f(x) + 2g(x))g'(x) \].

Step 5: Simplify the Expression

Now, simplify the expression:\[ \frac{dy}{dx} = f'(x)g(x) + 2g'(x)g(x) + f(x)g'(x) + 2g(x)g'(x) \].Combine like terms:\[ \frac{dy}{dx} = f'(x)g(x) + f(x)g'(x) + 4g(x)g'(x) \].

Key Concepts

Product Rule

The product rule is a differentiation technique used when you want to find the derivative of a product of two functions. Let's say you have two functions, \( u(x) \) and \( v(x) \). Instead of finding their derivatives separately, the product rule helps you differentiate their product, \( y = u(x)v(x) \), effectively. The product rule states that the derivative of \( y \) is given by:\[ \frac{dy}{dx} = u'(x) \cdot v(x) + u(x) \cdot v'(x) \]

1. Differentiate \( u(x) \): Find \( u'(x) \), the derivative of \( u \).
2. Differentiate \( v(x) \): Find \( v'(x) \), the derivative of \( v \).
3. Substitute: Substitute \( u'(x) \), \( v(x) \), \( u(x) \), and \( v'(x) \) back into the formula.

It's essential because it allows you to find derivatives where functions are multiplied together. This is useful in calculus when dealing with complex expressions.

Differentiable Functions

For a function to be differentiable, its derivative must exist at every point in its domain. This means that the function should be smooth without any sharp corners or discontinuities. When functions like \( f(x) \) and \( g(x) \) are differentiable, it simplifies the process of finding their derivatives, as seen in the original problem.Key characteristics of differentiable functions include:

• Continuity: The function must not have breaks, so it should be continuous.
• No sharp turns: Derivatives indicate the slope; hence the slope must change smoothly.
• Existence of Tangent Lines: At any point on the function, a tangent line can be drawn.

If \( f(x) \) and \( g(x) \) weren't differentiable, we couldn't easily apply differentiation techniques like the product rule, thus making calculus more complex.

Chain Rule

While the chain rule doesn't directly apply in the exercise, it's another differentiation technique closely related to differentiable functions. The chain rule helps differentiate composite functions, which are functions made by nesting one function inside another. If you have \( y = h(g(x)) \), the chain rule comes into play.The formula for the chain rule is:\[ \frac{dy}{dx} = h'(g(x)) \cdot g'(x) \]This rule is essential when a problem involves nested functions, allowing you to "chain" the derivatives together:

• Differentiate the outer function: Start with the outermost function \( h \).
• Multiply by the derivative of the inner function: Find the derivative of the inner function \( g \).

The chain rule ensures you efficiently handle layers of functions, crucial for advanced calculus problems.

Differentiation Technique

Differentiation is all about finding how a function changes — its rate of change. Different techniques have been developed for different types of functions, and they allow us to tackle diverse mathematical problems. Using the right technique makes solving these problems easier and more intuitive.

1. Product Rule

One technique is the product rule, used when dealing with a product of functions, useful in the given exercise.

2. Chain Rule

The chain rule is vital for differentiating composite functions. Even if not directly used, understanding it is key for more intricate problems.

3. Power Rule

This simple rule helps when differentiating functions of the form \( x^n \), with the derivative being \( nx^{n-1} \).Each differentiation technique serves particular scenarios, and mastering them allows for flexibility in calculus, helping to solve problems efficiently.