Problem 42 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ h(t)=4^{2^{3}-t} $$

Step-by-Step Solution

Answer

The derivative is \( -4^{8-t} \ln(4) \).

Step 1: Identify the Structure of the Function

The function given is \( h(t) = 4^{2^3 - t} \). This is an exponential function where the base is constant and the exponent involves the variable \( t \). To differentiate, we need to consider it in the context of the exponential function rule.

Step 2: Apply the Exponential Function Differentiation Rule

For a function in the form \( a^{f(x)} \), the derivative is \( a^{f(x)} \cdot f'(x) \cdot \, \ln(a) \). In our case, the base \( a \) is 4, and \( f(t) = 2^3 - t \). Thus, \( h(t)' = 4^{2^3 - t} \cdot (-1) \cdot \, \ln(4) \) since \( f'(t) = -1 \).

Step 3: Simplify the Expression

Calculate \( 2^3 \), which equals 8, so the function becomes \( 4^{8-t} \). Therefore, \( h(t)' = -4^{8-t} \cdot \, \ln(4) \).

Step 4: Conclude the Derivative

The final expression for the derivative of \( h(t) \) with respect to \( t \) is \( -4^{8-t} \ln(4) \).

Key Concepts

Exponential Functions

Exponential functions are mathematical expressions where the variable appears in the exponent. These functions have the general form \( a^{x} \) where \( a \) is a constant base and \( x \) is the exponent featuring the variable. The function \( h(t) = 4^{2^3 - t} \) is an example of an exponential function with a constant base of 4 and a variable exponent \( (2^3 - t) \). This structure makes exponential functions unique, as they grow rapidly compared to polynomial functions as the value of the exponent increases.
This fast-paced growth makes exponential functions incredibly useful in modeling scenarios involving growth, decay, or any process that intensifies over time, such as population growth, radioactive decay, and compounding interest in finance. In the context of differentiation, understanding how exponential functions behave is key to determining how their rates of change are influenced by the introduction of variables in the exponent.

Derivative Rules

Differentiating exponential functions involves specific rules. For a function \( a^{f(x)} \), the derivative is given by \( a^{f(x)} \cdot f'(x) \cdot \ln(a) \). Here, \( f(x) \) is the function in the exponent, \( f'(x) \) is its derivative, and \( \ln(a) \) is the natural logarithm of the base.

• Base : Constant, such as 4 in \( 4^{2^3 - t} \).
• Exponent: Involving the variable, such as \( 2^3 - t \).
• Derivative of exponent: Calculated as \( -1 \) when \( f(t) = 2^3 - t \).

The differentiation process for \( h(t) = 4^{8-t} \) uses these rules, resulting in \( h(t)' = -4^{8-t} \cdot \ln(4) \). This expression tells us not only how \( h(t) \) changes with \( t \) but also integrates the constant growth rate factor \( \ln(4) \). These steps and rules guide us in computing the rate of change of more complex expressions, providing a clear procedure for handling exponential functions.

Calculus

Calculus fundamentally explores change and motion through derivatives and integrals. One must understand the concept of the derivative: a fundamental idea capturing how a quantity changes as its input changes. This is often interpreted as the slope of the function's curve at any given point.
In the context of \( h(t) = 4^{2^3 - t} \), calculus helps us deduce how the function's growth rate alters over time. By differentiating this function, we determine an expression \( -4^{8-t} \ln(4) \), which explicitly conveys the function's instantaneous rate of change at any point \( t \). Such an expression is valuable in predicting and understanding dynamic systems, whether it’s for calculating financial trends or understanding natural phenomena.
In calculus, exponential functions often model real-world problems, requiring precise calculations of their growth rates or slopes—these derivatives help unravel deeper insights into the specific phenomena being studied, making calculus a crucial tool in both theoretical and applied sciences.