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Question
If `x = sqrt(a^(sin^(-1)))`, y = `sqrt(a^(cos^(-1)))` show that `dy/dx = - y/x`
Solution 1
The given equations are `x = sqrt(a^(sin^(-1)))`, y = `sqrt(a^(cos^(-1)t))`
Solution 2
Given: `x = sqrt(a^(sin^(-1)t))` and `y = sqrt (a^(cos^(-1)t))`
`dx/dt = 1/2 . 1/(a^(sin^(-1)t)). d/dt a ^(sin^(-1)t)`
`= 1/2 . 1/ sqrt (a^(sin^(-1)t)). a^(sin^(-1)t) . log a d/dt sin^-1 t`
`= sqrt((a^(sin^(-1)))t)/2. log a . 1/ (sqrt(1-t^2)`
`dy/dt = 1/2. 1/ sqrt (a^(cos^(-1))t). d/dt a^(cos^(-1)t)`
`= 1/2 . 1/sqrt (a^(cos^(-1))t). a^( cos^(-1)) . log a. -1/(sqrt (1 - t^2))`
`= sqrt (a^(cos^(-1))t)/2 .log a -1/sqrt(1 - t^2)`
`∴ dy/dx = (dy/dt)/(dx/dt) = (sqrt (a^(cos^(-1))t)/2. log a. -1/ sqrt(1 - t^2))/( sqrt (a^(sin^(-1))t)/2 . log a . 1/ sqrt (1 - t^2))`
`= (-sqrt( a^(cos^(-1)))t)/ sqrt (a^(sin^(-1))t) = (-y)/x.`
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