Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12034961

Differentiate the following w.r.t.x: square root of fraction numerator 1 minus sin space straight x over denominator 1 plus sin space straight x end fraction end root.

Solution
Let space space space space space straight y equals square root of fraction numerator 1 minus sin space straight x over denominator 1 plus sin space straight x end fraction end root equals square root of fraction numerator 1 minus sin space straight x over denominator 1 plus sin space straight x end fraction cross times fraction numerator 1 minus sin space straight x over denominator 1 minus sin space straight x end fraction end root
space space space space space space space space space space space space space equals square root of fraction numerator left parenthesis 1 minus sin space straight x right parenthesis squared over denominator 1 plus sin squared space straight x end fraction end root equals square root of fraction numerator left parenthesis 1 minus sin space straight x right parenthesis squared over denominator cos squared straight x end fraction end root equals fraction numerator 1 minus sin space straight x over denominator cos space straight x end fraction equals fraction numerator 1 over denominator cos space straight x end fraction minus fraction numerator sin space straight x over denominator cos space straight x end fraction equals sec space straight x minus tan space straight x
therefore space dy over dx equals sec space straight x space tan space straight x minus sec squared straight x equals sec space straight x left parenthesis tan space straight x minus sec space straight x right parenthesis

Some More Questions From Continuity and Differentiability Chapter