Problem 28 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(t)=\left(4 t^{4}+\frac{4}{t^{4}}\right)^{1 / 4}\)

Step-by-Step Solution

Answer

The derivative \(h'(t)\) is \(\frac{1}{4}(4t^4 + \frac{4}{t^4})^{-3/4}(16t^3 - \frac{16}{t^5})\).

Step 1: Identify the Function Type

The function provided is \(h(t)=\left(4 t^{4}+\frac{4}{t^{4}}\right)^{1 / 4}\). This is a composite function because it involves the outer function \((x)^{1/4}\) and the inner function \(x = 4t^4 + \frac{4}{t^4}\).

Step 2: Apply the Chain Rule

To differentiate the function, use the chain rule which states \(\frac{d}{dt}[f(g(t))] = f'(g(t)) \cdot g'(t)\). Here, \(f(x) = x^{1/4}\) and \(g(t) = 4t^4 + \frac{4}{t^4}\).

Step 3: Differentiate the Outer Function

Differentiate the outer function \(f(x) = x^{1/4}\). The derivative is \(f'(x) = \frac{1}{4}x^{-3/4}\).

Step 4: Differentiate the Inner Function

Differentiate the inner function \(g(t) = 4t^4 + \frac{4}{t^4}\). This gives us \(g'(t) = 16t^3 - \frac{16}{t^5}\).

Step 5: Combine Results Using the Chain Rule

Substitute \(g(t)\) and \(g'(t)\) back into the chain rule formula: \(h'(t) = f'(g(t)) \cdot g'(t) = \frac{1}{4}(4t^4 + \frac{4}{t^4})^{-3/4} \cdot (16t^3 - \frac{16}{t^5})\).

Step 6: Simplify the Expression

Combine the terms to simplify the expression: \[h'(t) = \frac{1}{4}(4t^4 + \frac{4}{t^4})^{-3/4} \cdot (16t^3 - \frac{16}{t^5})\]. Simplifying further is complex without numerical calculation, but conceptually this is the derivative.

Key Concepts

Composite Function

A composite function is essentially a function within a function. It is when you have something like the function \(f(g(t))\). Here, you take an output from one function and plug it into another. This kind of structure is common and a core part of calculus, particularly when you're dealing with complicated expressions.

In our example, the function given can be split into two parts:

• Outer function: \(f(x) = x^{1/4}\)
• Inner function: \(g(t) = 4t^4 + \frac{4}{t^4}\)

These two parts together compose the actual function \(h(t)\).

Understanding composite functions is crucial when differentiating complex functions. The innermost function, \(g(t)\), defines a new expression which is then transformed by the outer function, \(f\). This layered approach is what makes composite functions both intriguing and challenging.

Differentiation

Differentiation is a fundamental concept in calculus that refers to finding the derivative of a function. The process is akin to discovering the slope of a curve at a given point, or essentially, how fast a function is changing at that moment.

When you differentiate a function, you're looking to find its rate of change with respect to its independent variable. It involves applying specific rules and techniques to achieve this goal. For composite functions, this often means employing the chain rule—a strategy tailored for differentiating complex, layered functions.

In the exercise, applying differentiation requires us to perform only the chain rule, because we are dealing with both an inner and outer function. Differentiation turns into a step-by-step process as you peel back each layer, evaluating each component's change, then putting all the pieces back together.

Derivative

A derivative represents the rate at which a function is changing. It is a vital tool in calculus as it offers insights into the behavior of functions. The concept of a derivative forms the bedrock for more advanced topics like optimization and motion analysis.

The derivative of a function can be thought of as the slope of the tangent line to the graph of the function at a particular point. It's about capturing the essence of change and providing a mathematical framework to predict and analyze these changes.

• The outer function \(f(x) = x^{1/4}\) differentiates to \(f'(x) = \frac{1}{4}x^{-3/4}\).
• The inner function \(g(t) = 4t^4 + \frac{4}{t^4}\) yields \(g'(t) = 16t^3 - \frac{16}{t^5}\).

In our task, the derivative \(h'(t)\) is found by combining these results through the chain rule. The derivative, \(h'(t)\), signifies how \(h(t)\) changes with respect to \(t\), and is expressed as a combination of these two components, each describing a part of the function's structure.