If f(x) = cot^{-1}left(frac{x^x - x^{-x}}{2}r | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$.Let $u = x^x$. Then $f(x) = \cot^{-1}\left(\frac{u - u^{-1}}{2}\right) = \cot^{-1}\lef

Vedclass · Accepted Answer

-$1$

Vedclass · Answer

(A) Given $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}ight)$.Let $u = x^x$. Then $f(x) = \cot^{-1}\left(\frac{u - u^{-1}}{2}ight) = \cot^{-1}\left(\frac{u^2 - 1}{2u}ight)$.Using the identity $\cot^{-1}(z) = 	an^{-1}(1/z)$,we have $f(x) = 	an^{-1}\left(\frac{2u}{u^2 - 1}ight)$.Recall that $2	an^{-1}(u) = 	an^{-1}\left(\frac{2u}{1 - u^2}ight)$,so $f(x) = -2	an^{-1}(u) + \frac{\pi}{2}$ (for appropriate range).Differentiating with respect to $x$: $f'(x) = -2 \cdot \frac{1}{1 + u^2} \cdot \frac{du}{dx}$.Since $u = x^x$,$\frac{du}{dx} = x^x(1 + \ln x)$.At $x = 1$,$u = 1^1 = 1$ and $\frac{du}{dx} = 1^1(1 + \ln 1) = 1$.Thus,$f'(1) = -2 \cdot \frac{1}{1 + 1^2} \cdot 1 = -2 \cdot \frac{1}{2} = -1$.

If $f(x) = \cot^{-1}\left(\frac{x^x - x^{-x}}{2}\right)$,then the value of $f'(1)$ is equal to

Similar Questions

The differential coefficient of $f[\log (x)]$ with respect to $x$ when $f(x) = \log x$ is:

If $f(x)=3^x$ and $g(x)=4^x$,then $\frac{f^{\prime}(0)-g^{\prime}(0)}{1+f^{\prime}(0) g^{\prime}(0)}$ is

If $y = \log_{2026}(\log_{2025} x)$, then $\frac{dy}{dx} = \dots \dots \dots$

$\frac{d}{dx} \left( \log \sqrt{\frac{1+\sin x}{1-\sin x}} \right) = $

$\frac{d}{dx} \left[ \log \sqrt{\sin \sqrt{e^x}} \right] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module