Problem 27 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(t)=\left(3 t+\frac{3}{t}\right)^{2 / 5}\)

Step-by-Step Solution

Answer

The derivative of the function is \( \frac{2}{5} (3t + \frac{3}{t})^{-3/5} \cdot \left(3 - \frac{3}{t^2}\right) \).

Step 1: Identify the Composition of Functions

The function \( h(t) = (3t + \frac{3}{t})^{2/5} \) is a composition of functions. It can be seen as an outer function \( u^{2/5} \) where \( u = 3t + \frac{3}{t} \). We will apply the chain rule to differentiate this function.

Step 2: Differentiate the Outer Function

Using the chain rule, differentiate the outer function \( u^{2/5} \) with respect to \( u \). The derivative is \( \frac{d}{du} (u^{2/5}) = \frac{2}{5} u^{-3/5} \).

Step 3: Differentiate the Inner Function

The inner function is \( u = 3t + \frac{3}{t} \). The derivative with respect to \( t \) is \( \frac{du}{dt} = 3 - \frac{3}{t^2} \).

Step 4: Apply the Chain Rule

Combine the results from Steps 2 and 3 using the chain rule. This gives us the derivative of the composition: \( \frac{dh}{dt} = \frac{d}{du} (u^{2/5}) \cdot \frac{du}{dt} = \frac{2}{5} (3t + \frac{3}{t})^{-3/5} \cdot \left(3 - \frac{3}{t^2}\right) \).

Step 5: Simplify the Expression

The derivative expression can be simplified as \( \frac{dh}{dt} = \frac{2}{5} (3t + \frac{3}{t})^{-3/5} \cdot \left(3 - \frac{3}{t^2}\right) \). There's no further simplification needed unless specified for a particular form.

Key Concepts

Composition of Functions

In calculus and algebra, understanding the composition of functions is crucial. When functions are combined to form new functions, they are said to be 'composed' of one another, which is essential in solving many calculus problems.
The original exercise example of the function \(h(t)=(3t+\frac{3}{t})^{2/5}\) is an excellent representation of this concept. The function can be viewed as being made up of two parts: an outer function, represented by \(u^{2/5}\), and an inner function, \(u = 3t + \frac{3}{t}\).
When understanding or solving such problems, always identify these components:

• The **inner function** is what you apply first (\(3t + \frac{3}{t}\)).
• The **outer function** is applied to the result of the inner (\(u^{2/5}\)).

Recognizing this structure allows you to use the chain rule effectively to differentiate complex functions.

Derivatives

Derivatives are a foundational tool in calculus, providing the rate at which a function changes at any given point. This concept is essential for solving many real-world problems, from physics to economics.
To differentiate a composed function like \(h(t)\), you need to understand the derivatives of both the outer and inner functions separately:

• The **derivative of the outer function** \(u^{2/5}\) with respect to \(u\) is \(\frac{2}{5} u^{-3/5}\).
• The **derivative of the inner function** \(u = 3t + \frac{3}{t}\) with respect to \(t\) is \(3 - \frac{3}{t^2}\).

By calculating these derivatives, you can successfully apply them in the chain rule to find the overall derivative of the composition, efficiently solving the problem.

Calculus Problem Solving

The essence of calculus problem solving lies in correctly applying concepts such as the chain rule for differentiation. It's not just about knowing the formulas but understanding how and why they work.
In the given example, the chain rule is applied to differentiate the composed function \(h(t)=(3t+\frac{3}{t})^{2/5}\):

• First, compute the derivative of the outer function with respect to the inner.
• Then, find the derivative of the inner function with respect to the variable \(t\).
• Lastly, multiply these derivatives to obtain \(\frac{dh}{dt} = \frac{2}{5} (3t + \frac{3}{t})^{-3/5} \left(3 - \frac{3}{t^2}\right)\).

Breaking down the problem into these steps allows you to tackle even the complex expressions methodically.
Such a structured process not only helps in solving the problem at hand but also reinforces your understanding of calculus, making future problems easier to approach.