Question

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: The third derivative of the given function is $$f'''(x) = \frac{2}{x}$$.

Step-by-step solution

Find the first derivative of the function\(f(x) = x^2 \ln x\)

To begin, we can observe that the function is the product of two separate functions: $$u(x) = x^2$$ and $$v(x) = \ln x$$. In order to find the first derivative of f(x), we will use the product rule: $$(u \cdot v)' = u' \cdot v + u \cdot v'$$. First, let us find the derivatives of the functions u(x) and v(x): $$u'(x) = \frac{d}{dx}(x^2) = 2x$$ $$v'(x) = \frac{d}{dx}(\ln(x)) = \frac{1}{x}$$. Now, we apply the product rule: $$f'(x) = 2x \cdot \ln{x} + x^2 \cdot \frac{1}{x}$$ $$\implies 2x\ln{x} + x$$ So the first derivative is: $$f'(x) = 2x\ln{x} + x$$.

Find the second derivative of the function \(f'(x) = 2x\ln{x} + x\)

Again, we notice that the first term is a product of two functions, so we will use the product rule. Let the two functions for the first term be: $$u(x) = 2x$$ and $$v(x) = \ln x$$. Derivative of these functions are: $$u'(x) = \frac{d}{dx}(2x) = 2$$ $$v'(x) = \frac{d}{dx}(\ln{x}) = \frac{1}{x}$$. Now, the second term's derivative is straightforward: $$\frac{d}{dx}x = 1$$. Applying the product rule to the first term and combining the results: $$f''(x) = (2 \cdot \ln{x}) + (2x \cdot \frac{1}{x}) + 1$$ $$\implies 2\ln{x} + 2 + 1$$ So the second derivative is: $$f''(x) = 2\ln{x} + 3$$.

Find the third derivative of the function \(f''(x) = 2\ln{x} + 3\)

For the third derivative, we again apply the chain rule to the first term: $$\frac{d}{dx}(2\ln{x}) = 2\cdot \frac{d}{dx}\ln{x} = 2\cdot \frac{1}{x} = \frac{2}{x}$$ Now, the second term's derivative is zero, as it is a constant: $$\frac{d}{dx}(3) = 0$$. So, the third derivative is: $$f'''(x) = \frac{2}{x} + 0$$ Finally, the third derivative of the function is: $$\frac{d^3}{dx^3}(x^2 \ln x) = \frac{2}{x}$$.

Key concepts

Product Rule

When we are faced with a function that is the product of two functions, like in our example where the function is given by the expression f(x) = x^2 \times \text{ln}(x), we need a special technique to find the derivative. This technique is called the product rule.

The product rule can be summarized as 'the derivative of the first function times the second plus the first function times the derivative of the second.' In mathematical terms, if we have two functions u(x) and v(x), the product rule states that
\[ (u \times v)' = u' \times v + u \times v' \.\]
In our exercise, while applying this rule to find the first derivative, we got f'(x) = 2x \times \text{ln}(x) + x. It's crucial to differentiate both functions separately and then apply the formula. Understanding the product rule is fundamental in calculus, particularly when dealing with products of functions.

Chain Rule

The chain rule is a fundamental and powerful tool in calculus that helps us differentiate the composition of two or more functions. It essentially tells us to take the derivative of the outer function, multiplied by the derivative of the inner function. In the given example, when finding the higher-order derivatives, particularly the third derivative, we applied the chain rule to the logarithmic part of the function.

The basic form of the chain rule is:
\[ \frac{d}{dx}[g(h(x))] = g'(h(x)) \times h'(x) \.\]
When we applied the chain rule to the term 2\text{ln}(x) in our third step of the solution, we obtained the third derivative as f'''(x) = 2/x. Recognizing when to use the chain rule is essential for handling complex derivatives which involve nested functions.

Logarithmic Differentiation

Logarithmic differentiation is a method used to differentiate functions that are difficult to differentiate using the standard rules of differentiation, like the product or quotient rule. This technique is especially useful when dealing with variables in both the base and the exponent of a function, or for products and quotients of functions raised to powers.

The general method for logarithmic differentiation involves taking the natural logarithm (ln) of both sides of the equation y = f(x), then using the properties of logarithms to simplify. This is followed by differentiating implicitly with respect to x, and then solving for y'. Within the context of our higher-order derivatives problem, logarithmic differentiation could be used as an alternative approach to simplify the differentiation process for products or quotients of functions.

Calculus

Calculus is a vast field of mathematics focused on change and motion, encompassing concepts such as derivatives, integrals, and limits. It allows us to understand how quantities change with respect to each other and to find properties such as slopes, areas, and volumes. In our textbook solution example, we are dealing with higher-order derivatives, which are a core concept in calculus.

Higher-order derivatives refer to the subsequent derivatives taken after the first derivative, indicating how the rate of change itself is changing. For instance, if the first derivative represents velocity, then the second derivative — the derivative of the velocity — represents acceleration. In our exercise, by finding the first, second, and third derivatives, we are exploring deeper into the function's behavior and seeing how its slope changes at various points. It's important to grasp calculus concepts to solve real-world problems in science, engineering, economics, and beyond.