Problem 47 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(t)=5^{\sqrt{t}}\)

Step-by-Step Solution

Answer

The derivative of the function is \( h'(t) = \frac{5^{\sqrt{t}} \cdot \ln(5)}{2\sqrt{t}} \).

Step 1: Identify Function Type

The given function is \( h(t) = 5^{\sqrt{t}} \). This is an exponential function where the base is a constant and the exponent is a function of \( t \).

Step 2: Use Exponential Differentiation Rule

For functions of the form \( y = a^{u(t)} \), the derivative with respect to \( t \) is \( y' = a^{u(t)} \cdot \ln(a) \cdot \frac{d}{dt}(u(t)) \). In this case, \( a = 5 \) and \( u(t) = \sqrt{t} \).

Step 3: Differentiate the Exponent

Differentiate \( u(t) = \sqrt{t} = t^{1/2} \) with respect to \( t \). Use the power rule, which states that \( \frac{d}{dt}(t^n) = n \cdot t^{n-1} \). Thus, \( \frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}} \).

Step 4: Apply the Differentiation Formula

Substitute \( a = 5 \), \( u(t) = \sqrt{t} \), \( \ln(a) = \ln(5) \), and \( \frac{d}{dt}(u(t)) = \frac{1}{2\sqrt{t}} \) into the formula from Step 2: \[ h'(t) = 5^{\sqrt{t}} \cdot \ln(5) \cdot \frac{1}{2\sqrt{t}} \].

Step 5: Simplify the Expression

Combine the terms in the expression: \[ h'(t) = \frac{5^{\sqrt{t}} \cdot \ln(5)}{2\sqrt{t}} \].

Key Concepts

Understanding Exponential Functions

An exponential function is a mathematical expression where a constant base is raised to the power of a variable, often seen in forms like \( y = a^x \). These functions exhibit exponential growth or decay, depending on the value of the base. When the base is greater than one, the function grows, while a base between 0 and 1 leads to decay.

In our exercise, \( h(t) = 5^{\sqrt{t}} \) is an exponential function. Here, the base \( a = 5 \) is a constant and the exponent is the square root of \( t \), an expression which itself can change as \( t \) varies. This function illustrates how even complex expressions can form the exponent in exponential functions. Understanding the structure of these functions is key to solving calculus problems involving differentiation of such expressions.

Applying the Power Rule

The power rule is one of the foundational tools in calculus, used to differentiate expressions involving powers of a variable. For a function \( f(x) = x^n \), where \( n \) is any real number, the derivative is given as \( f'(x) = n \cdot x^{n-1} \). This rule simplifies the process of finding rates of change and tangents to curves.

In our differentiation example, we applied the power rule to \( u(t) = \sqrt{t} = t^{1/2} \). Using the power rule, we find that \( \frac{d}{dt}(t^{1/2}) = \frac{1}{2}t^{-1/2} = \frac{1}{2\sqrt{t}} \). This derivative is a crucial step in finding the overall derivative of the original exponential function.

The Process of Calculus Differentiation

Calculus differentiation is the process of finding the derivative, or rate of change, of a function. It involves understanding and applying various rules, such as the product rule, chain rule, and, importantly for this case, the rule for differentiating exponential functions.

For functions in the form \( y = a^{u(x)} \), the differentiation involves:

• Taking the original exponential function \( y = a^{u(t)} \).
• Using the differentiation formula: \( y' = a^{u(t)} \cdot \ln(a) \cdot \frac{d}{dt}(u(t)) \).
• Calculating \( \ln(a) \), the natural logarithm of the base.
• Finding the derivative of \( u(t) \), which could involve rules like the power rule.

For our example, this results in \( h'(t) = \frac{5^{\sqrt{t}} \cdot \ln(5)}{2\sqrt{t}} \). Mastering these steps helps in tackling more complex calculations in calculus.