Question

Find \(f^{\prime}(x)\) for each function. $$ f(x)=e^{x^{3} \ln x} $$

Short answer

\( f^{\prime}(x) = x^2 e^{x^3 \ln x} (3 \ln x + 1) \)

Step-by-step solution

Identify the Inner Functions

The given function is complex and can be expressed in terms of inner functions using a combination of exponential and natural logarithm functions. Identify the inner functions: let \( u(x) = x^3 \ln x \) such that the original function becomes \( f(x) = e^{u(x)} \).

Differentiate the Inner Function \(u(x)\)

Differentiate \( u(x) = x^3 \ln x \) using the product rule. \[ \frac{d}{dx}[x^3 \ln x] = \frac{d}{dx}[x^3] \cdot \ln x + x^3 \cdot \frac{d}{dx}[\ln x] \]Calculate each derivative: \( \frac{d}{dx}[x^3] = 3x^2 \) and \( \frac{d}{dx}[\ln x] = \frac{1}{x} \).Combine them: \[ 3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2 \ln x + x^2 \]

Differentiate the Outer Function

The outer function is an exponential function \( f(x) = e^{u(x)} \). Use the chain rule to find the derivative: \[ f^{\prime}(x) = e^{u(x)} \cdot \frac{d}{dx}[u(x)] \]Substitute \( \frac{d}{dx}[u(x)] = 3x^2 \ln x + x^2 \) from Step 2.Thus, \[ f^{\prime}(x) = e^{x^3 \ln x} \cdot (3x^2 \ln x + x^2) \]

Simplify the Derivative Expression

Simplify the expression for \( f^{\prime}(x) \).The derivative is: \[ f^{\prime}(x) = e^{x^3 \ln x} (3x^2 \ln x + x^2) \]This can be factored further into: \[ f^{\prime}(x) = x^2 e^{x^3 \ln x} (3 \ln x + 1) \]

Key concepts

Product Rule

When differentiating functions, it's essential to know when to use the product rule. If you have two functions multiplied together, their derivative is not just the derivative of each function multiplied. Instead, the product rule tells us:

• Calculate the derivative of the first function and keep the second function as it is.
• Keep the first function as it is, and calculate the derivative of the second function.

Finally, add these two results. In mathematical terms, for two functions \( u(x) \) and \( v(x) \), the derivative is given by:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]In this exercise, we used the product rule to differentiate \( u(x) = x^3 \ln x \). Here, \( x^3 \) is one function, and \( \ln x \) is another. Applying the product rule helps us find:\[3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2 \ln x + x^2\]Using the product rule correctly is crucial for functions where multiplication is involved.

Chain Rule

The chain rule is your go-to tool when you need to differentiate a composite function. Composite functions are functions within functions, like \( f(x) = e^{u(x)} \), where \( u(x) \) is itself another function.
To apply the chain rule, follow these steps:

• Differentiate the outer function, leaving the inner function unchanged.
• Multiply this result by the derivative of the inner function.

In this problem, the outer function is \( e^{u(x)} \), and the inner function is \( u(x) = x^3 \ln x \). The chain rule formula would thus be:\[f^{\prime}(x) = e^{u(x)} \cdot \frac{d}{dx}[u(x)]\] Inserting the value from our earlier derivative of \( u(x) \), we find:\[e^{x^3 \ln x} \cdot (3x^2 \ln x + x^2)\]Mastering the chain rule allows you to break down complex functions into simpler parts, making differentiation manageable.

Natural Logarithm Differentiation

The natural logarithm, denoted as \( \ln x \), is a common function you'll encounter in calculus. Differentiating \( \ln x \) has a straightforward rule: its derivative is \( \frac{1}{x} \).
This rule simplifies the process of dealing with logarithmic functions in differentiation. For instance, in the given exercise, we deal with a product \( x^3 \ln x \), and differentiating \( \ln x \) directly gives us \( \frac{1}{x} \).
The natural logarithm has unique properties that make it especially useful for simplifying expressions involving exponents. This exercise showcases how logarithmic differentiation can be combined with other rules, like the product and chain rules, to solve complex derivatives.

• Remember, the derivative of \( \ln x \) is always \( \frac{1}{x} \).

When dealing with problems involving natural logs, apply this rule wherever \( \ln x \) appears, and proceed with any additional rules as necessary. This understanding enriches your toolkit for handling exponential and logarithmic functions in calculus.