Problem 46 - Calculus of a Single Variable

Question

In Exercises 41–64, find the derivative of the function. $$ y=x^{2} \ln x $$

Step-by-Step Solution

Answer

The derivative of the function \(y = x^{2} \ln x \) is \( y' = x + 2x \ln x \).

Step 1: Identify the Functions for Product Rule

Firstly, identify the two functions which are multiplied together to make \(y\). Here, function \(u(x) = x^2\) and function \(v(x) = \ln x\). The Product Rule will be used to find the derivative.

Step 2: Compute the Derivatives of Both Functions

Now we need to find the derivatives of each function. The derivative of \(u(x)\) using the power rule is \(u'(x) = 2x\). The derivative of \(v(x) = \ln x\) is \(v'(x) = \frac{1}{x}\).

Step 3: Apply the Product Rule

According to the Product Rule, the derivative \(y'\) of \(y = u(x)v(x)\) is given by: \(y' = u(x)v'(x) + v(x)u'(x)\). Substituting our original functions and their derivatives, we get \( y' = x^2 * \frac{1}{x} + \ln x * 2x = x + 2x \ln x \).

Key Concepts

Derivative of a Function

Understanding the derivative of a function is essential in calculus. It represents the rate at which a function's output value is changing at any given point. In simpler terms, if you're driving a car, the speedometer tells you your instantaneous speed – the derivative tells us the equivalent in regard to functions.

Mathematically, if we have a function, let's say, \(f(x)\), its derivative, denoted as \(f'(x)\) or \(\frac{df}{dx}\), gives the slope of the tangent to the curve of \(f(x)\) at any point \(x\). This concept allows us to understand how one quantity varies with respect to another and is foundational in all of calculus.

Product Rule for Differentiation

When faced with the task of differentiating a product of two functions, just like in the example \(y = x^{2} \ln x\), we use the product rule for differentiation. This rule states that if you have two differentiable functions \(u(x)\) and \(v(x)\), the derivative of their product \(y = u(x)v(x)\) is \(y' = u'(x)v(x) + u(x)v'(x)\).

This rule is immensely useful as it systematically breaks down a complex expression into simpler parts that can be easily differentiated. It allows us to differentiate products without having to expand them, saving time and reducing the potential for errors.

Power Rule in Calculus

The power rule is one of the most straightforward differentiation rules, which is great news for students who are just starting out with calculus. It is used when differentiating functions of the form \(x^n\), where \(n\) is any real number.

According to the power rule, the derivative of \(x^n\) is \(nx^{n-1}\). This makes calculations quick and efficient. For example, in our exercise, the first part of the function was \(u(x) = x^2\), and using the power rule, its derivative was easily found to be \(u'(x) = 2x^(2-1) = 2x\). Understanding this rule not only simplifies many differentiation problems but also serves as a building block for more complex rules in calculus.

Logarithmic Differentiation

Logarithmic differentiation comes into play when dealing with products, quotients, or powers of functions where direct application of standard rules might be cumbersome. Although not used in the step-by-step solution earlier, it is an important concept to be familiar with.

It involves taking the natural logarithm of both sides of an equation before differentiating, which simplifies the differentiation process, especially when dealing with exponents and products. In our case, if we were to apply logarithmic differentiation, we would first take the natural logarithm of \(y = x^{2} \ln x\) and then use the properties of logarithms to simplify before differentiating. This process can sometimes provide an alternative method that could be more straightforward than traditional applications of the product or quotient rules.