InterviewSolution
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InterviewSolution
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निम्नलिखित में प्रत्येक का `(dy)/(dx)` ज्ञात कीजिए `y=tan^(-1){(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))}` |
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Answer» `y=tan^(-1){(sqrt(1+x^(2))-sqrt(1-x^(2)))/(sqrt(1+x^(2))+sqrt(1-x^(2)))}` माना `x^(2)=costheta` तब `theta=cos^(-1)x^(2)` `thereforey=tan^(-1){(sqrt(1+costheta)-sqrt(1-costheta))/(sqrt(1+costheta)+sqrt(1-costheta))}` `rArry=tan^(-1){(sqrt(2cos^(2)theta/2)-sqrt(2sin^(2)theta/2))/(sqrt(2cos^(2)theta/2)+sqrt(2sin^(2)theta/2))}` `rArry=tan^(-1){(costheta/2-sintheta/2)/(costheta/2+sintheta/2)}` `rArry=tan^(-1){(1-tantheta/2)/(1+tantheta/2)}` `rArry=tan^(-1){tan(pi/4-theta/2)}` `rArry=pi/4-theta/2` `rArry=pi/4-1/2cos^(-1)x^(2)` दोनों पक्षों का x के सापेक्ष अवकलन करने पर, `(dy)/(dx)=d/(dx){pi/4-1/2cos^(-1)x^(2)}` `rArr(dy)/(dx)=0-1/2(-1)/(sqrt(1-(x^(2))^(2)))d/(dx)(x^(2))` `rArr(dy)/(dx)=1/(2sqrt(1-x^(4)))xx2x` `=x/(sqrt(1-x^(4)))` |
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