Problem 47 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ h(t)=5^{\sqrt{t}} $$

Step-by-Step Solution

Answer

The derivative is \( \frac{dh}{dt} = 5^{\sqrt{t}} \cdot \frac{\ln{5}}{2\sqrt{t}} \).

Step 1: Understand the function

We are given the function \( h(t) = 5^{\sqrt{t}} \). Our task is to differentiate this with respect to \( t \). The function involves an exponent that is a function of \( t \), specifically \( \sqrt{t} \), which suggests the use of chain rule and possibly logarithmic differentiation.

Step 2: Apply logarithmic differentiation

To simplify differentiating \( h(t) = 5^{\sqrt{t}} \), take the natural logarithm of both sides, \( \ln{h(t)} = \ln{(5^{\sqrt{t}})} \). Simplifying, we have \( \ln{h(t)} = \sqrt{t} \ln{5} \) using the logarithm power rule \( \ln(a^b) = b \ln{a} \).

Step 3: Differentiate using implicit differentiation

To find \( \frac{dh}{dt} \), differentiate both sides of \( \ln{h(t)} = \sqrt{t} \ln{5} \) with respect to \( t \). The derivative of the left side is \( \frac{1}{h(t)} \frac{dh}{dt} \). For the right side, use the chain rule: \( \frac{d}{dt} (\sqrt{t} \ln{5}) = \frac{1}{2\sqrt{t}} \ln{5} \).

Step 4: Solve for \( \frac{dh}{dt} \)

We equate \( \frac{1}{h(t)} \frac{dh}{dt} \) to \( \frac{1}{2\sqrt{t}} \ln{5} \) from implicit differentiation. Thus, \( \frac{dh}{dt} = h(t) \cdot \frac{1}{2\sqrt{t}} \ln{5} \). Substitute \( h(t) = 5^{\sqrt{t}} \) back in, giving \( \frac{dh}{dt} = 5^{\sqrt{t}} \cdot \frac{\ln{5}}{2\sqrt{t}} \).

Key Concepts

Understanding the Chain Rule

The chain rule is a fundamental tool in calculus used to differentiate composite functions. The essence of the chain rule is that it helps you differentiate functions that are nested within each other.

Take, for instance, the function in your exercise: \( h(t) = 5^{\sqrt{t}} \). It involves an "inner function," \( u(t) = \sqrt{t} \), and an "outer function," \( v(u) = 5^u \).

To differentiate \( h(t) \) with respect to \( t \), you first differentiate the outer function with respect to the inner function, and then multiply by the derivative of the inner function with respect to \( t \).

• Differentiate the outer function: \( \frac{d}{du} (5^u) = 5^u \ln{5}\)
• Differentiate the inner function: \( \frac{d}{dt} (\sqrt{t}) = \frac{1}{2\sqrt{t}} \)
• Apply the chain rule: \( \frac{dh}{dt} = 5^{\sqrt{t}} \ln{5} \times \frac{1}{2\sqrt{t}} \)

This process highlights the idea of linking derivatives step-by-step. The chain rule simplifies the differentiation of composite functions, deeply rooted in many calculus problems.

Exploring Exponential Functions

Exponential functions form a core part of many mathematical concepts, characterized by having a constant base raised to a variable exponent. In the given exercise, \( h(t) = 5^{\sqrt{t}} \), \( 5 \) is the base, and \( \sqrt{t} \) is the exponent. Exponential functions showcase rapid growth or decay, depending on the sign and magnitude of the exponent.

They possess unique differentiation properties, where the rate of change of the function is proportional to the function itself, particularly noticeable in the natural exponential function \( e^x \). However, for other bases like \( 5 \), the differentiation involves an additional constant, the logarithm of the base:

• Differentiation: \( \frac{d}{dx}(a^x) = a^x \ln{a} \)

In our case:

• \( \frac{d}{du}(5^u) = 5^u \ln{5} \)

This formula is used when applying the chain rule to differentiate \( h(t) \), revealing exponential functions' simplicity and elegance.

Applying Logarithmic Differentiation

Logarithmic differentiation is a powerful technique for tackling functions with variable exponents and intricate products or quotients. By taking the logarithm of both sides, the differentiation process is greatly simplified.

For the function \( h(t) = 5^{\sqrt{t}} \), applying logarithmic differentiation involves the following steps:

• Start by taking the natural logarithm: \( \ln{h(t)} = \ln(5^{\sqrt{t}}) \)
• Use the logarithm power rule: \( \ln{h(t)} = \sqrt{t} \ln{5} \)
• Differentiate implicitly: differentiate both sides with respect to \( t \)

The left side becomes \( \frac{1}{h(t)}\frac{dh}{dt} \), and for the right side, apply the chain rule:

• \( \frac{d}{dt}(\sqrt{t} \ln{5}) = \frac{1}{2\sqrt{t}} \ln{5} \)

Finally, solve for \( \frac{dh}{dt} \) by isolating it:

• \( \frac{dh}{dt} = h(t) \cdot \frac{\ln{5}}{2\sqrt{t}} \)
• Substitute back \( h(t) = 5^{\sqrt{t}} \)
• \( \frac{dh}{dt} = 5^{\sqrt{t}} \cdot \frac{\ln{5}}{2\sqrt{t}} \)

This method is particularly useful when other differentiation techniques seem cumbersome, turning complex problems into manageable tasks.