Question

$If space straight y equals sin left parenthesis 2 space sin to the power of negative 1 end exponent straight x right parenthesis comma space show space that space dy over dx equals 2 square root of fraction numerator 1 minus straight y squared over denominator 1 minus straight x squared end fraction end root.$

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Solution

Here space space space space space space straight y equals sin left parenthesis 2 space sin to the power of negative 1 end exponent straight x right parenthesis Put space sin to the power of negative 1 end exponent straight x equals straight theta space straight i. straight e. space straight x equals sin space straight theta therefore space space space space space space space space space space space straight y equals sin space 2 straight theta space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis 1 right parenthesis therefore space space space space space dy over dx equals 2 space cos space 2 straight theta equals 2 square root of 1 minus sin squared space straight theta end root therefore space space space space space dy over dx equals 2 square root of 1 minus straight y squared end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space of space left parenthesis 1 right parenthesis right square bracket Now space dy over dx equals dy over dθ cross times dθ over dx space space space space space space space space space space space space space space space space space equals 2 square root of 1 minus straight y squared end root cross times fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction comma space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space straight theta equals sin to the power of negative 1 end exponent straight x right square bracket therefore space space space space space space dy over dx equals 2 square root of fraction numerator 1 minus straight y squared over denominator 1 minus straight x squared end fraction end root.

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