Q82 / 150 · MHT CET 2026 (18 April Shift 2) · Mathematics
For $x > 0$ and $(x \log x) < 1$, if $y = \cot^{-1}\left(\dfrac{x - \log x^{x^2}}{\log e^{x^2} + \log x^x}\right)$, then $\dfrac{dy}{dx} = \ldots$
- A$\dfrac{1}{1+x^2} + \dfrac{2}{x[1+(\log x)^2]}$
- Correct answerB$\dfrac{1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}$
- C$\dfrac{-1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}$
- D$\dfrac{1}{1+x^2} + \dfrac{1}{[1+(\log x)^2]}$
Solution
Given function is:
$y = \cot^{-1}\left(\dfrac{x - \log x^{x^2}}{\log e^{x^2} + \log x^x}\right)$
Simplify the numerator and the denominator of the argument:
Numerator $= x - x^2 \log x = x(1 - x \log x)$
Denominator $= x^2 \log e + x \log x = x^2 + x \log x = x(x + \log x)$
Substitute these back into the function:
$y = \cot^{-1}\left(\dfrac{x(1 - x \log x)}{x(x + \log x)}\right) = \cot^{-1}\left(\dfrac{1 - x \log x}{x + \log x}\right)$
We know the inverse trigonometric identity $\cot^{-1}\left(\dfrac{1 - ab}{a + b}\right) = \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right)$ (up to a constant multiple of $\pi$ depending on the sign of $a+b$, which becomes zero upon differentiation).
Using $a = x$ and $b = \log x$, and given $x \log x < 1$ (so $1 - ab > 0$):
$y = C + \tan^{-1}\left(\dfrac{x + \log x}{1 - x \log x}\right)$
$y = C + \tan^{-1} x + \tan^{-1}(\log x)$
Differentiating both sides with respect to $x$:
$\dfrac{dy}{dx} = \dfrac{d}{dx}(\tan^{-1} x) + \dfrac{d}{dx}(\tan^{-1}(\log x))$
$\dfrac{dy}{dx} = \dfrac{1}{1+x^2} + \dfrac{1}{1+(\log x)^2} \cdot \dfrac{d}{dx}(\log x)$
$\dfrac{dy}{dx} = \dfrac{1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}$
Answer: $\dfrac{1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}$
$y = \cot^{-1}\left(\dfrac{x - \log x^{x^2}}{\log e^{x^2} + \log x^x}\right)$
Simplify the numerator and the denominator of the argument:
Numerator $= x - x^2 \log x = x(1 - x \log x)$
Denominator $= x^2 \log e + x \log x = x^2 + x \log x = x(x + \log x)$
Substitute these back into the function:
$y = \cot^{-1}\left(\dfrac{x(1 - x \log x)}{x(x + \log x)}\right) = \cot^{-1}\left(\dfrac{1 - x \log x}{x + \log x}\right)$
We know the inverse trigonometric identity $\cot^{-1}\left(\dfrac{1 - ab}{a + b}\right) = \tan^{-1}\left(\dfrac{a + b}{1 - ab}\right)$ (up to a constant multiple of $\pi$ depending on the sign of $a+b$, which becomes zero upon differentiation).
Using $a = x$ and $b = \log x$, and given $x \log x < 1$ (so $1 - ab > 0$):
$y = C + \tan^{-1}\left(\dfrac{x + \log x}{1 - x \log x}\right)$
$y = C + \tan^{-1} x + \tan^{-1}(\log x)$
Differentiating both sides with respect to $x$:
$\dfrac{dy}{dx} = \dfrac{d}{dx}(\tan^{-1} x) + \dfrac{d}{dx}(\tan^{-1}(\log x))$
$\dfrac{dy}{dx} = \dfrac{1}{1+x^2} + \dfrac{1}{1+(\log x)^2} \cdot \dfrac{d}{dx}(\log x)$
$\dfrac{dy}{dx} = \dfrac{1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}$
Answer: $\dfrac{1}{1+x^2} + \dfrac{1}{x[1+(\log x)^2]}$
Hence, the correct answer is option: 'B'.