Problem 42 - Calculus for Biology and Medicine

Question

Differentiate the functions with respect to the independent variable. \(h(t)=4^{2 t^{3}-t}\)

Step-by-Step Solution

Answer

\( \frac{dh}{dt} = 4^{2t^3 - t} \cdot \ln(4) \cdot (6t^2 - 1) \)

Step 1: Identify the function and variable

The given function is \( h(t) = 4^{2t^3 - t} \) and the independent variable with respect to which we need to differentiate is \( t \).

Step 2: Apply the chain rule and exponential differentiation

We have an exponential function in the form \( a^{g(t)} \) where \( a = 4 \) and \( g(t) = 2t^3 - t \). The derivative of \( a^{g(t)} \) is given by \( a^{g(t)} \cdot \, \ln(a) \cdot \frac{d}{dt}[g(t)] \).

Step 3: Differentiate the inner function

Differentiate \( g(t) = 2t^3 - t \) with respect to \( t \). The derivative is \( \frac{d}{dt}[2t^3 - t] = 6t^2 - 1 \).

Step 4: Substitute and simplify

Substitute the derivatives back into the chain rule formula: \( \frac{dh}{dt} = 4^{2t^3 - t} \cdot \ln(4) \cdot (6t^2 - 1) \).

Step 5: Finalize the differentiated result

Thus, the derivative of the function \( h(t) = 4^{2t^3 - t} \) with respect to \( t \) is: \[ \frac{dh}{dt} = 4^{2t^3 - t} \cdot \ln(4) \cdot (6t^2 - 1) \].

Key Concepts

Chain Rule in Differentiation

The Chain Rule is a fundamental tool in calculus used for differentiating compositions of functions. Whenever you encounter a function nested inside another function, the Chain Rule is your go-to technique. To break it down:

• It helps to first identify the outer function and the inner function.
• You then differentiate the outer function with respect to the inner function.
• Afterward, you multiply that result by the derivative of the inner function itself.

For example, in the function given, the outer function can be thought of as an exponential function, and the inner function is a polynomial. By applying the Chain Rule, you efficiently find the derivative of the entire composition without unraveling it unnecessarily. This is invaluable for dealing with more complex calculus problems efficiently and accurately.

Understanding Exponential Functions

Exponential functions often appear in the form \( a^{x} \), where \( a \) is a constant base and \( x \) is the exponent, which could be a function itself. They are powerful because they describe situations involving growth or decay, like population growth or radioactive decay.
In differentiation, dealing with exponential functions requires extra steps compared to simpler polynomial forms. The key formula for the derivative of an exponential function \( a^{g(t)} \) is:
\[ a^{g(t)} \, \ln(a) \, \frac{d}{dt}[g(t)] \]
You multiply the original function by the natural logarithm of the base and the derivative of the exponent. This accounts for both the shape of the exponential curve and the rate at which the exponent function changes. By mastering this approach, exponential differentiation becomes a systematic process.

Differentiate Using Derivatives

Derivatives are a core concept in calculus, used to determine how a function changes as its input changes. The process of finding a derivative, known as differentiation, helps you understand the rate at which one quantity changes relative to another.
When you apply derivatives, you hone in on the slope of the tangent line to the curve at any given point, providing a precise measure of instantaneous change.

In the exercise with the function \( h(t)=4^{2t^3-t} \), you start with the inner polynomial function \( g(t) = 2t^3 - t \). Its derivative \( \frac{d}{dt}[2t^3-t] = 6t^2 - 1 \), as calculated, shows how the function \( g(t) \) changes with \( t \). By then applying this result within the greater derivative of the composite function, you gain a complete derivative that respects all the function's layers. This shows how derivatives unravel the complexity of functions into understandable, actionable insights.