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If `y= cot^(-1)(sqrt(1+x^2+1)/x) " then find " dy/dx` |
Answer» Let x=tan theta therefore `y=cot ^(-1)(sqrt((1+tan ^2 theta +1)/(tan theta)))` `=cot ^(-1)((sec theta+1)/(tan theta))` `=cot ^(-1)((1+cos theta)/(sin theta))` `cot ^-1 ((2cos^2""theta/(2))/(2sin""theta/2cos""(theta)/(2)))=cot^(-1)(cot""theta/2)` `=theta/2=1/2 tan ^(-1)x` `rArr dy /dx=1/2 d/dx tan ^(-1)=(1)/(2(1+x^2))` |
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