Problem 26 - Calculus and its applications

Question

Differentiate. $$ g(x)=-\frac{4}{5} e^{x^{3}} $$

Step-by-Step Solution

Answer

The derivative is \(-\frac{12}{5} x^2 e^{x^3}\).

Step 1: Identify the Function and Rewrite

The function given is \( g(x) = -\frac{4}{5} e^{x^{3}} \). Let's rewrite it as \( g(x) = -\frac{4}{5} e^{u} \) where \( u = x^3 \). This change will help in applying the chain rule easily.

Step 2: Differentiate the Exponential Function

The derivative of \( e^u \) with respect to \( u \) is \( e^u \). Since we have \( g(x) = -\frac{4}{5} e^{u} \), the derivative will be \( -\frac{4}{5} e^u \), with respect to \( u \).

Step 3: Apply the Chain Rule

Now apply the chain rule because \( u = x^3 \). The derivative of \( u \) with respect to \( x \) is \( \frac{d}{dx}(x^3) = 3x^2 \). Multiply the result from Step 2 by the derivative of \( u \). So, \( \frac{dg}{dx} = -\frac{4}{5} e^{x^3} \times 3x^2 \).

Step 4: Simplify the Expression

Simplify the expression for the derivative: \( \frac{dg}{dx} = -\frac{4}{5} \times 3x^2 \times e^{x^3} = -\frac{12}{5} x^2 e^{x^3} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that involves finding the rate at which a function changes. This rate of change is known as the derivative.
In simple terms, differentiation helps us understand how a function behaves at any given point.

• The derivative provides how fast or slow the function is changing.
• In practical terms, differentiation can tell us things like speed from position or the slope at a particular point on a curve.

In this exercise, we are tasked with differentiating a function that involves an exponential term. Differentiation rules, such as the chain rule and product rule, make this process systematic and manageable. Here, the chain rule was used because the function inside the exponential term depends on another variable raised to a power.

Exponential Function

An exponential function is a mathematical function of the form \( e^x \), where \(e\) is a constant approximately equal to 2.71828.
It describes situations where growth or decay of a quantity is proportional to its current value.

• Exponential functions are particularly useful in modeling real-world phenomena such as population growth, radioactive decay, and compound interest.
• The derivative of an exponential function, like \( e^{x^3} \), uses the base property where differentiation doesn't alter its nature, i.e., \( \frac{d}{dx} e^{u} = e^{u} \).

In our problem, the exponential function \( e^{x^3} \) represents the variable part of the function, helping us understand how it contributes to the overall behavior of the function \( g(x) = -\frac{4}{5} e^{x^3} \).

Calculus Tutorial

This problem leverages one of the core lessons often covered in a calculus tutorial: applying the chain rule during differentiation.
A calculus tutorial focusing on differentiation would typically explain techniques starting from basic rules to more advanced ones like the chain rule.

• The chain rule is essential when differentiating functions composites, like \( f(g(x)) \), where you differentiate the outer function with respect to the inner one and then multiply by the derivative of the inner function.
• It is especially useful when dealing with nested or layered functions, such as our function involving the exponential term.

In this exercise, we see the practical application of these concepts by breaking down a complex derivative into simpler parts using the chain rule, demonstrating one of the many techniques that make calculus a powerful mathematical tool.