Question

For \(x > 0\) and \((xlogx) < 1\), if \(y = cot^{-1}(\frac{x-logx^{x^2}}{loge^{x^2}+logx^x})\), then \(\frac{dy}{dx} = \ldots\)

• A. \(\frac{1}{1+x^2}+\frac{2}{x[1+(logx)^2]}\)
• B. \(\frac{1}{1+x^2}+\frac{1}{x[1+(logx)^2]}\)
• C. \(\frac{-1}{1+x^2}+\frac{1}{x[1+(logx)^2]}\)
• D. \(\frac{1}{1+x^2}+\frac{1}{[1+(logx)^2]}\)

Hint:

Simplify the logarithms first. The argument becomes (1 - x log x)/(x + log x), so y = tan^-1 x + tan^-1(log x).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
We first simplify the fraction inside \(\cot^{-1}\) using logarithm rules, then convert it into a sum of two inverse tangents, which are easy to differentiate.

Step 2: Key Formula or Approach:
1. \(\log x^{x^2} = x^2\log x\), \(\log e^{x^2} = x^2\), \(\log x^x = x\log x\).
2. \(\cot^{-1}u = \tan^{-1}\dfrac{1}{u}\) and \(\tan^{-1}a + \tan^{-1}b = \tan^{-1}\dfrac{a+b}{1-ab}\) when \(ab < 1\).

Step 3: Detailed Explanation:
\[ \frac{x - x^2\log x}{x^2 + x\log x} = \frac{x(1 - x\log x)}{x(x + \log x)} = \frac{1 - x\log x}{x + \log x} \]
So
\[ y = \cot^{-1}\frac{1 - x\log x}{x + \log x} = \tan^{-1}\frac{x + \log x}{1 - x\log x} \]
With \(a = x\), \(b = \log x\), and the condition \(x\log x < 1\), this is
\[ y = \tan^{-1}x + \tan^{-1}(\log x) \]
Differentiate:
\[ \frac{dy}{dx} = \frac{1}{1 + x^2} + \frac{1}{1 + (\log x)^2}\cdot\frac{1}{x} = \frac{1}{1+x^2} + \frac{1}{x[1 + (\log x)^2]} \]
Option (A) has an extra factor 2 on the second term, option (C) has a wrong sign on the first term, and option (D) forgets the \(1/x\) from the chain rule.

Final Answer:
\(\dfrac{dy}{dx} = \dfrac{1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}\), option (B). \[ \boxed{\frac{1}{1+x^2}+\frac{1}{x[1+(\log x)^2]} \text{ (B)}} \]