Problem 26 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(t)=\left(t^{4}-5 t\right)^{5 / 2}\)

Step-by-Step Solution

Answer

The derivative is \( h'(t) = \frac{5}{2} (t^4 - 5t)^{3/2} \cdot (4t^3 - 5) \).

Step 1: Recognize the Function Type

The function given is \[ h(t) = (t^4 - 5t)^{5/2} \]This is a composite function, where outer function is \[ u^{5/2} \]and inner function is \[ u = (t^4 - 5t) \].We should use the chain rule to differentiate it.

Step 2: Apply the Chain Rule

According to the chain rule, the derivative of a composite function \( h(t) = f(g(t)) \) is given by\[ h'(t) = f'(g(t)) \cdot g'(t) \].Here, let \[ u = g(t) = t^4 - 5t \]and \[ h(t) = f(u) = u^{5/2} \].We'll need to find both \( f'(u) \) and \( g'(t) \).

Step 3: Differentiate the Outer Function

Differentiate the outer function \( f(u) = u^{5/2} \) with respect to \( u \) using the power rule:\[ f'(u) = \frac{5}{2} u^{3/2} \].

Step 4: Differentiate the Inner Function

Differentiate the inner function \( g(t) = t^4 - 5t \) with respect to \( t \):\[ g'(t) = 4t^3 - 5 \].

Step 5: Substitute and Simplify

Substitute \( u = t^4 - 5t \) and the derivatives back into the chain rule formula:\[ h'(t) = \left( \frac{5}{2} (t^4 - 5t)^{3/2} \right) \cdot (4t^3 - 5) \].Simplify the expression to get the final derivative:\[ h'(t) = \frac{5}{2} (t^4 - 5t)^{3/2} \cdot (4t^3 - 5) \].

Key Concepts

Chain Rule

The chain rule is a fundamental rule in calculus used to differentiate composite functions. Suppose you have two functions, an outer function \( f \) and an inner function \( g \), combined to form a single composite function \( h(t) = f(g(t)) \). Differentiating such functions involves more than just applying the derivative to each part separately. Instead, the chain rule provides a method to handle these compositions.

When you apply the chain rule, the derivative is calculated as follows:

• First, differentiate the outer function with respect to the inner function, noted as \( f'(g(t)) \).
• Then, multiply it by the derivative of the inner function \( g(t) \) with respect to \( t \), noted as \( g'(t) \).

Thus, the derivative of the composite function \( h(t) = f(g(t)) \) is \( h'(t) = f'(g(t)) \cdot g'(t) \).
This process helps in tackling complex differentiation problems by breaking them into manageable parts, each more straightforward to handle. In the case of our original function \( h(t) = (t^4 - 5t)^{5/2} \), the chain rule simplifies the differentiation process significantly.

Composite Functions

Composite functions arise when one function is nested inside another. Mathematically, it is expressed as \( h(t) = f(g(t)) \), where \( f \) is the outer function and \( g \) is the inner function.
Understanding composite functions is essential, particularly in scenarios requiring the chain rule.
Take for example our exercise, where the function \( h(t) \) is composed of an outer power function \((t^4 - 5t)^{5/2}\) and the inner polynomial \(t^4 - 5t\).

• The inner function forms the core polynomial expression \( u = t^4 - 5t \).
• The outer function is the power function \( u^{5/2} \), acting on the polynomial expression.

To differentiate composite functions effectively, it's helpful to "separate" these functions conceptually. Then apply appropriate differentiation techniques like the chain rule and power rule to find their derivatives easily. Recognizing a function as a composite one allows you to systematically apply these techniques, facilitating a more structured approach to solving calculus problems. With practice, identifying and dealing with composite functions becomes second nature.

Power Rule

The power rule is a fundamental technique in calculus for differentiating expressions of the form \( u^n \). This rule states that if you have a function \( f(u) = u^n \), then its derivative is \( f'(u) = n \cdot u^{n-1} \).
This rule is incredibly useful due to its simplicity and applicability to many functions you’ll encounter. When dealing with composite functions, such as those that necessitate the chain rule, the power rule often comes into play for the outer function differentiation.

In our example, the outer function \( u^{5/2} \) relies on the power rule for its differentiation:

• Differentiating with respect to \( u \), we arrive at \( f'(u) = \frac{5}{2} \cdot u^{3/2} \).

Subsequently, this derivative is combined with the inner derivative as per the chain rule. The power rule simplifies the task of finding the derivative of these polynomial forms, making it an indispensable tool in any calculus toolkit. With its straightforward application, the power rule forms the backbone of differentiation strategies for both basic and composite functions.