If y = sin^{-1}left[cos sqrt{frac{1+x}{2}}rig | vedclass

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

Question

(C) Given $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight] + x^x$. Using $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$,we have: $y = \si

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) Given $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight] + x^x$. Using $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$,we have: $y = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - \sqrt{\frac{1+x}{2}}ight)ight] + x^x = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$. Differentiating with respect to $x$: $\frac{dy}{dx} = 0 - \frac{1}{\sqrt{2}} \cdot \frac{d}{dx}((1+x)^{1/2}) + \frac{d}{dx}(x^x)$. For $x^x$,let $u = x^x$,then $\ln u = x \ln x$. $\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$,so $\frac{du}{dx} = x^x(1 + \ln x)$. Thus,$\frac{dy}{dx} = -\frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + x^x(1 + \ln x)$. At $x = 1$: $\left[\frac{dy}{dx}ight]_{x=1} = -\frac{1}{2\sqrt{2}\sqrt{2}} + 1^1(1 + \ln 1) = -\frac{1}{4} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

If $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right] + x^x$,then find $\frac{dy}{dx}$ at $x = 1$.

Similar Questions

If $y = {\left( {1 + \frac{1}{x}} \right)^x}$,then $\frac{dy}{dx} = $

If $y = x^{\sqrt{x}}$,then $\frac{dy}{dx} = $

Differentiate the function with respect to $x$: $(\log x)^{\cos x}$

$y = (\tan x)^{(\tan x)^{\tan x}}$,then at $x = \frac{\pi}{4}$,the value of $\frac{dy}{dx} = $

Differentiate the function with respect to $x$: $(\log x)^{x}+x^{\log x}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module