Question

If \(y=x^3\log x, then\ \frac{d^2y}{dx^2}\) is:

• A. 5+6logx
• B. x(3+2logx)
• C. x(5+6logx)
• D. x²(3+2logx)

Answer 1

The Correct Option is C

To find the second derivative \(\frac{d^2y}{dx^2}\) for the function \(y = x^3 \log x\), we first need to calculate the first derivative. We use the product rule for differentiation which states if \(u(x) \cdot v(x)\), then the derivative is \(\frac{d}{dx}[u \cdot v] = u' \cdot v + u \cdot v'\).

Let \(u = x^3\) and \(v = \log x\). Then \(u' = 3x^2\) and \(v' = \frac{1}{x}\).

First derivative:

\(\frac{dy}{dx} = u'v + uv' = 3x^2 \log x + x^3 \cdot \frac{1}{x} = 3x^2 \log x + x^2\).

Now, find the second derivative by differentiating \(\frac{dy}{dx}\):

\(\frac{d^2y}{dx^2}\) requires differentiating each term separately.

1. Differentiate \(3x^2 \log x\) using product rule:

Let \(a = 3x^2\) and \(b = \log x\).

\(a' = 6x\), \(b' = \frac{1}{x}\).

\(\frac{d}{dx}(3x^2 \log x) = a'b + ab' = 6x \log x + \frac{3x^2}{x}\)

= \(6x \log x + 3x\).

2. Differentiate \(x^2\):

\(\frac{d}{dx}(x^2) = 2x\).

Combining these results, we have:

\(\frac{d^2y}{dx^2} = 6x \log x + 3x + 2x = 6x \log x + 5x\).

Factor out \(x\):

\(\frac{d^2y}{dx^2} = x(5 + 6\log x)\).

Thus, the correct answer is: \(x(5 + 6\log x)\)