मराठी

If x = sqrt(a^(sin^(-1)t)), y = sqrt(a^(cos^(-1)t)) show that dy/dx = - y/x. - Mathematics

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प्रश्न

If x = `sqrt(a^(sin^(-1)t))`, y = `sqrt(a^(cos^(-1)t))` show that `dy/dx = - y/x`.

बेरीज
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उत्तर

Given, x = `sqrt(a^(sin^(-1)t))` and y = `sqrt(a^(cos^(-1)t))`

`dx/dt = 1/2 . 1/(sqrt(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)t) . 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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पाठ 5: Continuity and Differentiability - Exercise 5.6 [पृष्ठ १८१]

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एनसीईआरटी Mathematics Part 1 and 2 [English] Class 12
पाठ 5 Continuity and Differentiability
Exercise 5.6 | Q 11 | पृष्ठ १८१

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