x के सापेक्ष फलन का अवकलन कीजिए: cot ^{-1}lef | vedclass

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

Question

(C) माना $y = \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$.हम जानते हैं कि $1+\sin x = \cos^2 \frac

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(C) माना $y = \cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$.हम जानते हैं कि $1+\sin x = \cos^2 \frac{x}{2} + \sin^2 \frac{x}{2} + 2\sin \frac{x}{2} \cos \frac{x}{2} = (\cos \frac{x}{2} + \sin \frac{x}{2})^2$.इसी प्रकार,$1-\sin x = (\cos \frac{x}{2} - \sin \frac{x}{2})^2$.चूंकि $0  \sin \frac{x}{2}$.अतः,$\sqrt{1+\sin x} = \cos \frac{x}{2} + \sin \frac{x}{2}$ और $\sqrt{1-\sin x} = \cos \frac{x}{2} - \sin \frac{x}{2}$.इन मानों को व्यंजक में रखने पर:$y = \cot ^{-1}\left[\frac{(\cos \frac{x}{2} + \sin \frac{x}{2}) + (\cos \frac{x}{2} - \sin \frac{x}{2})}{(\cos \frac{x}{2} + \sin \frac{x}{2}) - (\cos \frac{x}{2} - \sin \frac{x}{2})}\right]$$y = \cot ^{-1}\left[\frac{2\cos \frac{x}{2}}{2\sin \frac{x}{2}}\right]$$y = \cot ^{-1}(\cot \frac{x}{2}) = \frac{x}{2}$.अतः,$\frac{dy}{dx} = \frac{d}{dx}(\frac{x}{2}) = \frac{1}{2}$.

$x$ के सापेक्ष फलन का अवकलन कीजिए: $\cot ^{-1}\left[\frac{\sqrt{1+\sin x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right]$,जहाँ $0 < x < \frac{\pi}{2}$.

Similar Questions

यदि $y=\tan ^{-1}\left(\frac{5 x+1}{3-x-6 x^2}\right)$ है,तो $\frac{d y}{d x}=$

यदि $y = \frac{1}{\sqrt{a^2 - b^2}} \cos^{-1} \left[ \frac{a \cos(x - \alpha) + b}{a + b \cos(x - \alpha)} \right]$ है,तो $\frac{dy}{dx} = $

यदि $y = \tan^{-1}\left(\frac{4x}{1+5x^2}\right) + \tan^{-1}\left(\frac{3+8x}{8-3x}\right)$ है,तो $\frac{dy}{dx} = $

यदि $y=\tan ^{-1}\left(\frac{2+3 x}{3-2 x}\right)+\tan ^{-1}\left(\frac{4 x}{1+5 x^2}\right)$ है,तो $\frac{d y}{d x}=$

यदि $y=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^2}\right)}{\log \left(ex^2\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$ है,तो $\frac{d^2 y}{dx^2}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module