Question
यदि $y = \sec ({\tan ^{ - 1}}x),$ तब $\frac{{dy}}{{dx}} =$
$= \sec ({\tan ^{ - 1}}x) \tan ({\tan ^{ - 1}}x) \cdot \frac{{1}}{{1 + {x^2}}}$
$ = \frac{{x}}{{1 + {x^2}}}\,.\,\sqrt {1 + {x^2}} = \frac{{x}}{{\sqrt {1 + {x^2}} }}$, $({\tan ^{ - 1}}x = {\sec ^{ - 1}}\sqrt {1 + {x^2}} )$.
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$\sum_{x \in \mathbb{R}}\left(\sin \left(\left(x^2+x+5\right) \frac{\pi}{2}\right)-\cos \left(\left(x^2+x+5\right) \pi\right)\right)$ बरांबर है__________
$\tan ^{-1} \frac{1-x}{1+x}=\frac{1}{2} \tan ^{-1} x,(x>0)$