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Continuity And Differentiability

Question
CBSEENMA12035153

Find space dy over dx space if space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 5 straight x plus 12 square root of 1 minus straight x squared end root over denominator 13 end fraction close parentheses

Solution
Here space straight y equals sin to the power of negative 1 end exponent open parentheses fraction numerator 5 straight x plus 12 square root of 1 minus straight x squared end root over denominator 13 end fraction close parentheses equals sin to the power of negative 1 end exponent open parentheses 5 over 13 straight x plus 12 over 13 square root of 1 minus straight x squared end root close parentheses
space space space space space space space space space space space space equals sin to the power of negative 1 end exponent open square brackets straight x square root of 1 minus open parentheses 12 over 13 close parentheses squared end root plus 12 over 13 square root of 1 minus straight x squared end root close square brackets
therefore space space space space space space straight y equals sin to the power of negative 1 end exponent straight x plus sin to the power of negative 1 end exponent 12 over 13
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 open square brackets because space sin to the power of negative 1 end exponent straight theta plus sin to the power of negative 1 end exponent straight ϕ equals sin to the power of negative 1 end exponent open curly brackets straight theta square root of 1 minus straight ϕ squared end root plus straight ϕ square root of 1 minus straight x squared end root close curly brackets close square brackets
therefore space dy over dx equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction plus 0
therefore space dy over dx equals fraction numerator 1 over denominator square root of 1 minus straight x squared end root end fraction

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