Problem 44 - Calculus for Biology and Medicine

Question

Assume that \(f(x)\) and \(g(x)\) are differentiable at \(x\). Find an expression for the derivative of \(y .\) $$ y=[-2 f(x)-3 g(x)] g(x)+\frac{2 g(x)}{3} $$

Step-by-Step Solution

Answer

The derivative of \( y \) is \([-2f'(x) - 3g'(x)]g(x) + [-2f(x) - 3g(x)]g'(x) + \frac{2}{3}g'(x)\).

Step 1: Derivative of a Product

The expression for \( y \) includes \( [-2f(x) - 3g(x)]g(x) \), which is a product of two functions. Use the product rule, which states that the derivative of \( u(x)v(x) \) is \( u'(x)v(x) + u(x)v'(x) \). Here, \( u(x) = -2f(x) - 3g(x) \) and \( v(x) = g(x) \). Compute the derivative of each component: \( u'(x) = -2f'(x) - 3g'(x) \) and \( v'(x) = g'(x) \).

Step 2: Apply Product Rule

Apply the product rule: \[\frac{d}{dx}([-2f(x) - 3g(x)]g(x)) = (u'v) + (uv') = [-2f'(x) - 3g'(x)]g(x) + [-2f(x) - 3g(x)]g'(x)\]

Step 3: Derivative of a Constant Multiplication

Now consider the second term \( \frac{2g(x)}{3} \). The derivative of a constant times a function is the constant times the derivative of the function. Thus, the derivative is \( \frac{2}{3}g'(x) \).

Step 4: Combine Derivatives

Combine the results from Steps 2 and 3 to obtain the full derivative:\[\frac{dy}{dx} = [-2f'(x) - 3g'(x)]g(x) + [-2f(x) - 3g(x)]g'(x) + \frac{2}{3}g'(x)\]

Step 5: Simplify the Result

There are no like terms that can be combined further in the expression. The final expression for the derivative is:\[\frac{dy}{dx} = [-2f'(x) - 3g'(x)]g(x) + [-2f(x) - 3g(x)]g'(x) + \frac{2}{3}g'(x)\]

Key Concepts

Product Rule

When you have a product of two functions and want to find its derivative, you cannot simply take the derivative of each function separately and then multiply them. Instead, you need to use the product rule. This is because the derivative of a product, such as \(u(x) \, v(x)\), involves both the derivatives of each function and the original functions themselves.
The product rule is given by:

• If \(y = u(x) \, v(x)\), then the derivative of \(y\) is \(y' = u'(x) \, v(x) + u(x) \, v'(x)\).

This formula helps you find the rate of change in products of functions, a common situation in calculus. Using this rule ensures that both functions' rates of change contribute to the overall derivative.For example, in the exercise, the expression \([-2f(x) - 3g(x)]g(x)\) can be treated as the product of \(u(x) = -2f(x) - 3g(x)\) and \(v(x) = g(x)\). By applying the product rule, you break down the process to easily find the derivative step-by-step.

Differentiable Functions

A function is said to be differentiable at a point if it has a derivative at that point. This means you can find the instantaneous rate of change or the slope of the tangent line of the function at that particular point.
Differentiability is a key concept in calculus. It provides important information about the behavior of a function around given points. For instance, many functions you'll encounter are differentiable when their graphs are smooth and continuous with no sharp turns or corners. In the context of the exercise, both \(f(x)\) and \(g(x)\) are differentiable. This is critical because it ensures that their derivatives \(f'(x)\) and \(g'(x)\) exist and can be calculated. Knowing that these functions are differentiable allows us to apply derivatives reliably when using the product rule or constant multiplication rule.

Constant Multiplication Rule

The constant multiplication rule is a fundamental principle for finding derivatives when functions are multiplied by constants. It's straightforward: the derivative of a constant times a function is just the constant times the derivative of the function. This rule can be summarized as:

• If \(y = c \, f(x)\) where \(c\) is a constant, then the derivative \(y' = c \, f'(x)\).

In simpler terms, when modifying a function by a constant, its rate of change is similarly scaled by that same constant. For example, in our problem, the term \(\frac{2}{3}g(x)\) required us to find its derivative. Using the constant multiplication rule, you take the derivative of \(g(x)\) and multiply it by the constant: \((\frac{2}{3}) \, g'(x)\).
This rule is useful because it lets you quickly simplify and reduce terms during differentiation, making calculations more manageable and efficient.