Problem 100 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(U\) and \(V\) are two functions of \(x\) having derivatives of the \(n\)th order, then \((U V)_{n}=U_{n} V+{ }^{n} C_{1} U_{n-1} V_{1}+{ }^{n} C_{2} U_{n-2} V_{2}+\ldots\) \(+{ }^{n} C_{r} U_{n-r} V_{r}+\ldots+{ }^{n} C_{n} U V_{n}\) If \(I_{n}=\frac{d^{n}}{d x^{n}}\left(x^{n} \log x\right)\), then \(I_{n}=n I_{n-1}+k\), where \(k=\) (A) \(n !\) (B) \((n-1) !\) (C) \((n-2) !\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is (B) \((n-1)!\).

Step 1: Identify the Functions

To solve the problem, we must focus on differentiating the function given, which is \(x^n \log x\). This function can be split as \(U(x) = x^n\) and \(V(x) = \log x\).

Step 2: Understand the Derivative Order

We need to find \(I_n = \frac{d^n}{dx^n}(x^n \log x)\). According to the question, the result ends up being a combination of previous derivatives and a constant \(k\).

Step 3: Application of Leibniz Rule

We apply the generalized product rule (or Leibniz's rule) here for the n-th derivative: \((U V)_n\). This is expanded into a series of terms, each containing combinations of derivatives of \(U\) and \(V\).

Step 4: Differentiate Logarithmic Function

The derivatives of \(\log x\) are \(V_1 = \frac{1}{x}\), \(V_2 = -\frac{1}{x^2}\), etc. Since we need \(n\) derivatives, the highest order term will be \(V_n\).

Step 5: Differentiate Power Function

The derivatives of \(x^n\) are \(U_1 = n x^{n-1}\), \(U_2 = n(n-1)x^{n-2}\), etc., with the n-th derivative being \(U_n = n!\). This term appears first in the application of Leibniz's rule.

Step 6: Evaluate the Expression

Using the expansion, the first term is from the complete derivative of \(x^n\): \(n! \log x\). Each additional term involves lower derivatives of \(x^n\) and derivatives of \(\log x\), reducing degrees of \(x^n\) and increasing powers of \(-1/x^k\).

Step 7: Identify and Isolate Constants

The constant \(k\) emerges as we simplify: it comes from the complete disappearance of the logarithm term and the factorial coefficient from the constant \(\frac{d}{dx} (x^n)\). Notably, \(k = (n-1)!\) emerges from this procedure.

Step 8: Verify and Match Options

The given options include factorial forms. Matching the derived \(k = (n-1)!\) with the choices, it corresponds to option (B) \((n-1)!\).

Key Concepts

n-th Derivative

The n-th derivative refers to the process of differentiating a function repeatedly n times. It is an essential concept in calculus, especially when dealing with functions that are dependent on multiple variables or need higher-order derivatives to understand their behavior fully. The notation used for the n-th derivative of a function \( f(x) \) is \( f^{(n)}(x) \) or \( \frac{d^n}{dx^n} f(x) \).

When finding the n-th derivative of a product of two functions, such as in the given problem, it often involves using advanced techniques like Leibniz's rule. This requires us to calculate combinations of successive derivatives of each function. Understanding these interactions is crucial as they give rise to polynomial-like expressions that help unravel the complex patterns within higher derivatives.

In the context of the exercise, the key was to understand how to apply these repetitive derivatives systematically to arrive at a simplified expression for \( I_n = \frac{d^n}{dx^n}(x^n \log x) \). Every step in this process builds on the derivatives calculated before, aiding the understanding of patterns such as factorial growth in terms like \( n! \) in the expansions.

Product Rule

The product rule is a fundamental rule in calculus used to find the derivative of the product of two functions. It states that if you have two differentiable functions, \( U(x) \) and \( V(x) \), the derivative of their product is given by:
\[ (UV)' = U'V + UV' \]
This simple principle escalates to what is known as Leibniz's rule when considering the n-th derivative, particularly when differentiating higher-order terms like \( (UV)_n \). This expands into a summation of terms involving combinations of derivatives of the individual functions, as seen in the formula:

• \(U_n V + \binom{n}{1} U_{n-1} V_1 + \cdots + \binom{n}{r} U_{n-r} V_r + \cdots + U V_n\)

In our case, differentiating \( x^n \log x \) utilized this rule extensively. By acknowledging that each degradation of \( x \) in the expansion follows a predictable pattern through the binomial coefficients \( \binom{n}{r} \), a solid understanding of the role of the product rule simplifies tackling intricate derivatives. This systematic unfolding unveiled the necessary constants in the expression, ensuring accurate results.

Logarithmic Differentiation

Logarithmic differentiation is a handy technique when dealing with functions of the form \( x^n \log x \) or expressions where direct application of differentiation rules becomes cumbersome. The essence of this method lies in taking logarithms of both sides of the function and then differentiating it.

When you have a function like \( x^n \log x \), applying logarithmic differentiation can simplify the process by breaking down multiplicative functions into additive forms under the logarithm. Then, standard differentiation rules can be applied more straightforwardly. This method is particularly potent when functions with variables in both exponents and bases are involved.

• For example, differentiating \( \log(x^n) = n \log x \) simplifies integration with other terms.
• Similarly, \( x^n \log x \) presents complexities that become more manageable through such transformations.

Ultimately, as seen in our exercise, the natural log's derivative properties (\( \frac{d}{dx} \log x = \frac{1}{x} \)) helps organize the outcome of each differentiation step, culminating in easier isolation of terms like the constant \( k = (n-1)! \). This approach consistently yields structured derivative patterns, ensuring accuracy while simplifying calculations.