Question

If \(x \in [-1,1]\) and \(y=(\cot^{-1}x)^{\cot^{-1}x}\), then \(\left(\frac{dy}{dx}\right)_{x=0}=\)

• A. \(\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}\left(1+\log\frac{\pi}{2}\right)\)
• B. \(\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}\left(1-\log\frac{\pi}{2}\right)\)
• C. \(-\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}\left(1+\log\frac{\pi}{2}\right)\)
• D. \(\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}\left(\log\frac{\pi}{2}-1\right)\)

Hint:

For expressions of the form \(u^u\), always use logarithmic differentiation: \[ \ln y=u\ln u. \] This converts the exponential form into a product, making differentiation much easier.

Answer 1

The Correct Option is C

Concept: For functions of the form \(y=u^u\), logarithmic differentiation is the most convenient method. \[ \ln y=u\ln u \] Differentiating both sides gives \[ \frac{1}{y}\frac{dy}{dx} = u'(1+\ln u). \]

Step 1: Let \(u=\cot^{-1}x\).
Then \[ y=(\cot^{-1}x)^{\cot^{-1}x}. \] Taking logarithm on both sides, \[ \ln y=(\cot^{-1}x)\ln(\cot^{-1}x). \]

Step 2: Differentiate both sides.
Since \[ \frac{d}{dx}(\cot^{-1}x) = -\frac{1}{1+x^2}, \] we get \[ \frac{1}{y}\frac{dy}{dx} = -\frac{1}{1+x^2} \left(1+\ln(\cot^{-1}x)\right). \] Hence, \[ \frac{dy}{dx} = -(\cot^{-1}x)^{\cot^{-1}x} \cdot \frac{1+\ln(\cot^{-1}x)}{1+x^2}. \]

Step 3: Evaluate at \(x=0\).
Since \[ \cot^{-1}(0)=\frac{\pi}{2}, \] we have \[ y= \left(\frac{\pi}{2}\right)^{\frac{\pi}{2}}. \] Therefore, \[ \left(\frac{dy}{dx}\right)_{x=0} = -\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}} \left(1+\ln\frac{\pi}{2}\right). \] center minipage0.65

\( \left(\frac{dy}{dx}\right)_{x=0} = -\left(\frac{\pi}{2}\right)^{\frac{\pi}{2}} \left(1+\log\frac{\pi}{2}\right) \) minipage center