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

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(C) दिया गया है $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight] + x^x$. $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$ का उपयोग करने पर:

Vedclass · Accepted Answer

$\frac{3}{4}$

Vedclass · Answer

(C) दिया गया है $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}ight] + x^x$. $\cos 	heta = \sin\left(\frac{\pi}{2} - 	hetaight)$ का उपयोग करने पर: $y = \sin^{-1}\left[\sin\left(\frac{\pi}{2} - \sqrt{\frac{1+x}{2}}ight)ight] + x^x = \frac{\pi}{2} - \sqrt{\frac{1+x}{2}} + x^x$. $x$ के सापेक्ष अवकलन करने पर: $\frac{dy}{dx} = 0 - \frac{1}{\sqrt{2}} \cdot \frac{d}{dx}((1+x)^{1/2}) + \frac{d}{dx}(x^x)$. $x^x$ के लिए,मान लीजिए $u = x^x$,तो $\ln u = x \ln x$. $\frac{1}{u} \frac{du}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1$,इसलिए $\frac{du}{dx} = x^x(1 + \ln x)$. अतः,$\frac{dy}{dx} = -\frac{1}{\sqrt{2}} \cdot \frac{1}{2\sqrt{1+x}} + x^x(1 + \ln x)$. $x = 1$ पर: $\left[\frac{dy}{dx}ight]_{x=1} = -\frac{1}{2\sqrt{2}\sqrt{2}} + 1^1(1 + \ln 1) = -\frac{1}{4} + 1(1 + 0) = -\frac{1}{4} + 1 = \frac{3}{4}$.

यदि $y = \sin^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right] + x^x$ है,तो $x = 1$ पर $\frac{dy}{dx}$ ज्ञात कीजिए।

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