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

Step 1: Choose a smart substitution.
For $y=u^{u}$ where $u=\cot^{-1}x$, logarithmic differentiation is cleanest.
Step 2: Take logs.
\[ \ln y=u\ln u=(\cot^{-1}x)\ln(\cot^{-1}x). \]
Step 3: Differentiate both sides.
Using $\dfrac{d}{dx}\cot^{-1}x=-\dfrac{1}{1+x^2}$, \[ \frac{1}{y}\frac{dy}{dx}=-\frac{1}{1+x^2}\big(1+\ln(\cot^{-1}x)\big). \]
Step 4: Solve for the derivative.
\[ \frac{dy}{dx}=-(\cot^{-1}x)^{\cot^{-1}x}\cdot\frac{1+\ln(\cot^{-1}x)}{1+x^2}. \]
Step 5: Plug in $x=0$.
Here $\cot^{-1}0=\dfrac{\pi}{2}$ and $1+x^2=1$, so $y=\left(\dfrac{\pi}{2}\right)^{\pi/2}$.
Step 6: Write the value.
\[ \left(\frac{dy}{dx}\right)_{x=0}=-\left(\frac{\pi}{2}\right)^{\pi/2}\left(1+\ln\frac{\pi}{2}\right). \]
\[ \boxed{-\left(\frac{\pi}{2}\right)^{\pi/2}\left(1+\log\frac{\pi}{2}\right)} \]