Problem 40 - Calculus for Biology and Medicine

Question

Differentiate the functions in Problems 1-52 with respect to the independent variable. $$ f(x)=3^{x^{3}-1} $$

Step-by-Step Solution

Answer

The derivative is \( 3x^2 \ln(3) \cdot 3^{x^3 - 1} \).

Step 1: Identify the Components

The function given is \( f(x) = 3^{x^3 - 1} \). Here, the base is 3, which is a constant, and the exponent is a function of \( x \), specifically \( x^3 - 1 \). We need to differentiate this function with respect to \( x \).

Step 2: Apply the Exponential Differentiation Rule

For a function of the form \( a^{u(x)} \), its derivative can be found using the formula \( \frac{d}{dx}[a^{u(x)}] = a^{u(x)} \ln(a) \cdot u'(x) \). Here, \( a = 3 \) and \( u(x) = x^3 - 1 \).

Step 3: Differentiate the Exponent

Calculate \( u'(x) \), the derivative of the exponent \( x^3 - 1 \). The derivative of \( x^3 \) is \( 3x^2 \) and the derivative of a constant \( -1 \) is 0. Thus, \( u'(x) = 3x^2 \).

Step 4: Plug into the Differentiation Formula

Substitute \( a = 3 \), \( u(x) = x^3 - 1 \), and \( u'(x) = 3x^2 \) into the formula from Step 2:\[ \frac{d}{dx}[3^{x^3 - 1}] = 3^{x^3 - 1} \ln(3) \cdot 3x^2 \]

Step 5: Simplify the Expression

Combine the terms to simplify the expression:\[ \frac{d}{dx}[3^{x^3 - 1}] = 3^{x^3 - 1} \cdot \ln(3) \cdot 3x^2 = 3x^2 \ln(3) \cdot 3^{x^3 - 1} \]

Key Concepts

Derivative of Exponential Functions

Exponential functions have a constant base raised to a variable exponent. Differentiating these functions involves using special rules because the operation involves both the base and the exponent. In the case of exponentials, let's consider a function like \( f(x) = a^{u(x)} \), where \( a \) is a constant base and \( u(x) \) is an exponent depending on \( x \). The derivative of an exponential function is identified by the rule:\[ \frac{d}{dx}[a^{u(x)}] = a^{u(x)} \ln(a) \cdot u'(x) \]This rule tells us that:

• You first keep the original exponential function \( a^{u(x)} \).
• Then, multiply it by the natural logarithm of the base, \( \ln(a) \).
• Finally, multiply by the derivative of the exponent \( u'(x) \).

Understanding this rule is crucial for solving problems that involve exponential growth or decay in real-world scenarios, such as population dynamics or radioactive decay.

Chain Rule in Calculus

The chain rule is an essential principle in calculus, used to differentiate compositions of functions. When you have a function composed inside another function, the chain rule allows you to differentiate it by observing how the outer and inner functions interact.If you have a function like \( f(g(x)) \), the derivative is found by:\[ \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \]Here's how you apply it:

• First, differentiate the outer function \( f \) with respect to its inner function \( g(x) \).
• Second, multiply by the derivative of the inner function \( g(x) \).

In exponential differentiation, this rule is used when the exponent itself is a function, like \( u(x) \) in \( a^{u(x)} \). This requires finding \( u'(x) \) first, using the power rule or other differentiation techniques, and then applying the chain rule to finalize the derivative of the overall function.

Power Rule for Derivatives

The power rule is one of the simplest and most commonly used rules in differentiation. It provides an easy way to find the derivative of a function of the form \( x^n \), where \( n \) is any real number. According to the power rule:\[ \frac{d}{dx}[x^n] = n \cdot x^{n-1} \]Here’s how you use it:

• You multiply the exponent \( n \) by the front of the expression.
• Reduce the exponent by one, changing \( x^n \) to \( x^{n-1} \).

For example, in the derivative of \( x^3 \), apply the power rule to get \( 3x^2 \). This method simplifies finding \( u'(x) \) when the exponent \( u(x) \) in exponential functions like \( 3^{x^3 - 1} \) includes terms like \( x^3 \). The power rule helps break down the exponent differentiation step by step, allowing the application of other rules like the exponential derivative and chain rule seamlessly.