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यदि (If) `x=sqrt(asin^(-1))t,y=sqrt(a^(cos-1)t)` दिखाएँ कि (show that ) `(dy)/(dx)=-(y)/(x)`

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दिया है, `x=sqrt(a^(sin^(-1))t)` … (1)
तथा `y=sqrt(a^(cos^(-1))t)` … (2)
`x^(2)y^(2)=a^(sin^(-1)t).a^(cos-1_(t))=a^(sin^(-1_(t)))cos^(-1)t=a^(pi/2)`
`[becausesin^(-1)t+cos^(-1)t=(pi)/(2)]`
इस प्रकार `x^(x)y^(2)=a^(pi/2)` या `xy=a^(pi/4)` ... (3)
दोनों पक्षों को x के सापेक्ष अवकलित करने पर हमें मिलता है,
`1.y+x(dy)/(dx)=0therefore(dy)/(dx)=-(y)/(x)`

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