If ` y=x^(n)log nx.,then (d^(2)y)/(dx^(2))=`

Options

A. ` x^(n-2) (n-1-n (n-1) log nx )`
B. ` x^(n-2) (n-1+ n (n-1) log nx )`
C. ` x^(n-2) (2n-1- n (n-1) log nx )`
D. ` x^(n-2) (2n-1+ n (n-1) log nx )`

Answer

To solve the problem \( y = x^n \log(nx) \) and find the second derivative \( \frac{d^2y}{dx^2} \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Using the product rule, we have:

\[ \frac{dy}{dx} = \frac{d}{dx}(x^n) \cdot \log(nx) + x^n \cdot \frac{d}{dx}(\log(nx)) \]

Calculating each part:

1. The derivative of \( x^n \) is \( nx^{n-1} \).
2. The derivative of \( \log(nx) \) using the chain rule is \( \frac{1}{nx} \cdot n = \frac{1}{x} \).

Putting it all together:

\[ \frac{dy}{dx} = x^n \cdot \frac{1}{x} + \log(nx) \cdot nx^{n-1} = x^{n-1} + nx^{n-1} \log(nx) \]

Now, we can factor out \( x^{n-1} \):

\[ \frac{dy}{dx} = x^{n-1} \left( 1 + n \log(nx) \right) \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left( x^{n-1} \left( 1 + n \log(nx) \right) \right) \]

Using the product rule again:

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(x^{n-1}) \cdot \left( 1 + n \log(nx) \right) + x^{n-1} \cdot \frac{d}{dx}\left(1 + n \log(nx)\right) \]

Calculating each part:

1. The derivative of \( x^{n-1} \) is \( (n-1)x^{n-2} \).
2. The derivative of \( 1 + n \log(nx) \) is \( n \cdot \frac{1}{x} = \frac{n}{x} \).

Putting it all together:

\[ \frac{d^2y}{dx^2} = (n-1)x^{n-2} \left( 1 + n \log(nx) \right) + x^{n-1} \cdot \frac{n}{x} \]

This simplifies to:

\[ \frac{d^2y}{dx^2} = (n-1)x^{n-2} \left( 1 + n \log(nx) \right) + n x^{n-2} \]

Now, we can combine the terms:

\[ \frac{d^2y}{dx^2} = x^{n-2} \left( (n-1)(1 + n \log(nx)) + n \right) \]

Expanding this gives:

\[ \frac{d^2y}{dx^2} = x^{n-2} \left( n + n(n-1) \log(nx) \right) \]

### Final Result

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = x^{n-2} \left( 2n - 1 + n(n-1) \log(nx) \right) \]