Question

$$ \text { } , \text { find the indicated derivative. } $$ $$ D_{x} e^{x^{3} \ln x} $$

Short answer

\( D_x e^{x^{3} \ln x} = e^{x^{3} \ln x} (3x^2 \ln x + x^2) \).

Step-by-step solution

Recall the Chain Rule

To find the derivative of a composite function, use the chain rule. The chain rule states that if you have a function composition \( f(g(x)) \), then its derivative is \( f'(g(x))g'(x) \).

Identify the Outer and Inner Functions

In the given function \( e^{x^{3} \, \ln x} \), the outer function is \( e^u \) where \( u = x^{3} \ln x \). This means we need to find \( e^u \)'s derivative with respect to \( u \) as part of our chain rule application.

Differentiate the Exponential Function

Differentiate the outer function. The derivative of \( e^u \) with respect to \( u \) is simply \( e^u \). So, part of our derivative is \( e^{x^{3} \, \ln x} \).

Differentiate the Inner Function

Now, differentiate the inner function \( x^{3} \ln x \). Use the product rule which states that for two functions \( u(x)v(x) \), the derivative is \( u'(x)v(x) + u(x)v'(x) \). Here, \( u(x) = x^3 \) and \( v(x) = \ln x \).

Apply the Product Rule

Take the derivatives: \( u'(x) = 3x^2 \) and \( v'(x) = \frac{1}{x} \). Applying the product rule gives us: \( 3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2 \ln x + x^2 \).

Combine Using Chain Rule

Using the chain rule, multiply the derivative of the outer function by the derivative of the inner function: \( e^{x^{3} \ln x} (3x^2 \ln x + x^2) \).

Simplify the Expression

Distribute the exponential function into the expression: \( e^{x^{3} \, \ln x} \cdot (3x^2 \ln x + x^2) \). This is the derivative of the given function with respect to \( x \).

Key concepts

Chain Rule

The Chain Rule is an essential tool in calculus, particularly when dealing with composite functions. To understand it, think of a composite function as a "function within a function." In mathematical terms, this is written as \( f(g(x)) \). The Chain Rule provides a way to differentiate such functions by breaking them down into their outer and inner components.

When applying the Chain Rule, you first find the derivative of the outer function while keeping the inner function unchanged. Next, you multiply this result by the derivative of the inner function.

• Formula: If you have \( f(g(x)) \), the derivative is \( f'(g(x))g'(x) \).

This rule is especially useful because it allows the differentiation of more complex functions systematically. Understanding and applying the Chain Rule correctly is critical for finding derivatives in more advanced calculus problems.

Exponential Function Differentiation

Differentiating exponential functions is straightforward and relies on one key principle: the derivative of \( e^u \) with respect to \( u \) is \( e^u \). This property makes calculations involving the natural exponential function simpler when compared to other functions, such as polynomials or trigonometric functions.

Whenever you differentiate an exponential function, remember this rule.

• Basic principle: The derivative of \( e^x \) is itself, \( e^x \).

It's particularly useful in combination with the Chain Rule, where the derivative of the outer function \( e^u \) is needed. This straightforward derivative pairs perfectly with the more complex inner derivatives, ensuring computations remain manageable even with nested expressions.

Product Rule

The Product Rule is another fundamental differentiation technique. It is used when you need to find the derivative of a product of two functions. If you have two functions \( u(x) \) and \( v(x) \), their product is \( u(x)v(x) \). Here's how the Product Rule works:

• For the product of two functions \( u(x) \) and \( v(x) \), the derivative is \( u'(x)v(x) + u(x)v'(x) \).

In our original exercise, the Product Rule is crucial for differentiating the inner function \( x^3 \ln x \). You treat \( u(x) = x^3 \) and \( v(x) = \ln x \), and apply the rule to find their combined derivative.

Using the Product Rule allows you to systematically break down and differentiate functions that consist of multiple terms multiplied together.

Inner and Outer Functions

Understanding the distinction between inner and outer functions is key to correctly applying the Chain Rule. In a composite function, the inner function is the core that is plugged into the outer function. Practically, this means identifying the two tiers of your function setup.

For example, consider the function \( e^{x^3 \ln x} \). Here:

• The outer function is \( e^u \), which indicates how the inner function is transformed.
• The inner function is \( x^3 \ln x \), which forms the core variable expression.

Knowing which function holds which position is crucial when calculating derivatives, especially with the Chain Rule in play. First, you work with the differentiation of the outer function, followed by the derivative of the inner function, before combining them as prescribed by the Chain Rule formula.