Question

Higher-order derivatives Find the following higher-order derivatives. $$\frac{d^{3}}{d x^{3}}\left(x^{2} \ln x\right)$$

Short answer

Question: Find the third derivative of the function $$f(x) = x^{2} \ln x$$. Answer: $$f^{(3)}(x) = 16x\ln x + 16x$$.

Step-by-step solution

Differentiate the function with respect to x for the first time

To differentiate the function \(f(x) = x^2 \ln x\) with respect to x, we will use the product rule of differentiation, which states that \((uv)' = u'v + uv'\), where \(u\) and \(v\) are functions of x. So, let \(u = x^2\) and \(v = \ln x\). Now differentiate both functions with respect to x: $$u' = \frac{d}{dx}(x^{2}) = 2x$$ $$v' = \frac{d}{dx}(\ln x) = \frac{1}{x}$$ Now apply the product rule: $$f'(x) = u'v + uv' = \left(2x\right)\left(\ln x\right) + \left(x^2\right)\left(\frac{1}{x}\right) = 2x\ln x + x$$

Differentiate the function with respect to x for the second time

Next, we will differentiate \(f'(x) = 2x\ln x + x\) with respect to x. Let \(u = 2x\ln x\) and \(v = x\). Now differentiate both functions with respect to x: $$u' = \frac{d}{dx}(2x\ln x) = 2\ln x + 2$$ $$v' = \frac{d}{dx}(x) = 1$$ Now apply the product rule: $$f''(x) = u'v + uv' = \left(2\ln x + 2\right)\left(x\right) + \left(2x\ln x + x\right)\left(1\right) = 2x\ln x + 4x$$

Differentiate the function with respect to x for the third time

Finally, we will differentiate \(f''(x) = 2x\ln x + 4x\) with respect to x. Let \(u = 2x\ln x\) and \(v = 4x\). Now differentiate both functions with respect to x: $$u' = \frac{d}{dx}(2x\ln x) = 2\ln x + 2$$ $$v' = \frac{d}{dx}(4x) = 4$$ Now apply the product rule: $$f^{(3)}(x) = u'v + uv' = \left(2\ln x + 2\right)\left(4x\right) + \left(2x\ln x\right)\left(4\right) = 8x\ln x + 16x + 8x\ln x = 16x\ln x + 16x$$ The third derivative of the function is $$f^{(3)}(x) = 16x\ln x + 16x$$.

Key concepts

Product Rule in Differentiation

When dealing with multiplication of two functions in calculus, the product rule is a key concept. It's essential for finding the derivative of a product of two functions. The product rule formula is \[(uv)' = u'v + uv'\]where \(u\) and \(v\) are both functions of \(x\). To apply this rule, you differentiate each function individually and multiply them as shown in the formula.
The original example uses the product rule several times, as the expression inside requires several derivatives. Notice how each multiplication step features the derivative of one function combined with the other function held constant, and then vice versa. Being proficient in this rule is crucial, as it appears frequently in higher-order derivatives.

Understanding Differentiation

Differentiation is the process of finding the derivative of a function, which measures how a function's output changes as its input changes. The derivative gives us the slope of the function at any given point. The process of differentiating involves applying various rules, of which the product rule is just one example.
In the example provided, differentiation is used repeatedly to find higher-order derivatives. Each new derivative is found by differentiating the result of the previous step, which systematically applies differentiation rules. This helps in understanding the rate of change at higher levels. Becoming comfortable with these rules is essential for moving forward in calculus.

Role of Natural Logarithm in Differentiation

The natural logarithm, denoted as \(\ln x\), is a function that frequently appears in calculus. When differentiating a logarithmic function, we apply specific derivative rules. The derivative of \(\ln x\) is \[\frac{d}{dx}(\ln x) = \frac{1}{x}\].This function is often paired with other functions, like polynomials, to form products.

• Recognize that the appearance of \(\ln x\) complicates the differentiation slightly due to its unique properties.
• Ensure proper use of log rules once combined with others, particularly in product rule scenarios.

In the provided exercise, you see \(\ln x\) handled through the product rule, enabling you to derive successive derivatives without unnecessary complexity.

Applying the Chain Rule

The chain rule is an essential formula for handling compositions of functions. Unlike the product rule, it’s used when you have a function inside another function, and it's crucial for finding derivatives in such cases. The chain rule states: \[(f(g(x)))' = f'(g(x)) \cdot g'(x)\].It is not directly applied in the provided solution, but knowing when and how to use it, especially regarding nested functions, is important.

• Use the chain rule when differentiated variables are interdependent.
• Whenever a function inside another function requires differentiation, the chain rule simplifies this task.

While the exercise at hand leverages mostly the product rule, familiarity with the chain rule helps grasp the interplay of functions in complex derivatives as you advance.