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Continuity And Differentiability

Question
CBSEENMA12035155
Wired Faculty App

If space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses comma space find space dy over dx.

Solution
straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses space space equals cos to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses plus sin to the power of negative 1 end exponent open parentheses fraction numerator straight x plus 1 over denominator straight x minus 1 end fraction close parentheses 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 sec to the power of negative 1 end exponent straight x equals cos to the power of negative 1 end exponent open parentheses 1 over straight x close parentheses close square brackets space space equals straight pi over 2 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 open square brackets because space sin to the power of negative 1 end exponent straight theta plus cos to the power of negative 1 end exponent straight theta equals straight pi over 2 close square brackets therefore space dy over dx equals 0

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