Question

In Exercises, find the derivative of the function. $$ f(x)=x \ln e^{x^{2}} $$

Short answer

\(f'(x) = 3x^2\)

Step-by-step solution

Recognize the forms

We first recognize that the function \(f(x) = x \ln (e^{x^2})\) is in the product form, \(u(x) v(x)\), where \(u(x) = x\) and \(v(x) = \ln (e^{x^2})\), and that \(v(x) = \ln (e^{x^2})\) is a composition of functions that can be simplified using properties of the natural logarithm.

Simplify the function using logarithmic properties

We can use the property of logarithms that says \(\ln a^b = b \ln a\) to simplify the function \(v(x)\). Therefore, our function \(f(x)\) becomes \(f(x) = x \cdot (x^2)\) equivalent to \(f(x) = x^3\).

Apply the power rule

The power rule states that the derivative of \(x^n\) is \(nx^{n-1}\). Therefore, the derivative \(f'(x)\) of our function \(f(x) = x^3\) is \(f'(x) = 3x^2\).

Key concepts

Product Rule

In calculus, the product rule is a helpful technique when you need to find the derivative of a function that is the product of two simpler functions. Think of it as a way to "break down" complex problems. Suppose you have a function that looks like this:

• \(f(x) = u(x) \cdot v(x)\)

where \(u(x)\) and \(v(x)\) are both functions of \(x\). To find the derivative of this function, you can follow the product rule, which is:

• \(f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\)

Here’s a quick breakdown:

• First, take the derivative of \(u(x)\), called \(u'(x)\), and multiply it by \(v(x)\).
• Then, take \(u(x)\) and multiply it by the derivative of \(v(x)\), called \(v'(x)\).
• Finally, add these two results together to find \(f'(x)\), the derivative of the original function.

In this specific exercise, recognizing that the original function \(f(x) = x \ln(e^{x^2})\) can be split into \(u(x) = x\) and \(v(x) = \ln(e^{x^2})\) is key. You'll see that simplifying \(v(x)\) can make applying the product rule more straightforward.

Logarithmic Properties

Logarithmic properties are powerful tools that simplify complex expressions involving logs. In the context of calculus, these properties can ease the process of differentiation. Here’s a handy property used often:

• \( \ln(a^b) = b \ln(a) \)

This tells us that taking the logarithm of a power means you can bring the exponent down in front as a multiplier. It's particularly useful when dealing with compositions of functions. Let's take a closer look at the function in the exercise:

• \( v(x) = \ln(e^{x^2}) \)

Using the property \(\ln(a^b) = b \ln(a)\), \(v(x)\) simplifies to \(x^2 \cdot \ln(e)\). Since \(\ln(e) = 1\), this simplifies further to \(x^2\).

• Thus, \(\ln(e^{x^2}) = x^2\).

By simplifying \(v(x)\) using logarithmic properties, we can transform complicated expressions into something we can easily differentiate or work with further.

Power Rule

The power rule is one of the simplest and most frequently used rules when differentiating functions in calculus. It states:

• The derivative of \(x^n\) with respect to \(x\) is \(nx^{n-1}\).

Here’s how it works:

• To differentiate \(x^3\), use the power rule by bringing the exponent, 3, to the front as a multiplier.
• Then subtract one from the exponent, making it \(3-1 = 2\).

Applying to our simplified function from the exercise, \(f(x) = x^3\):

• The derivative \(f'(x) = 3x^2\).

The power rule makes it quick and easy to find derivatives of polynomial expressions like this. Once you grasp the logic, it's applicable to any power of \(x\). Remember, this process is for functions where the base is \(x\), and the exponent is a constant number, making it highly versatile in calculus problems.