+91-8076753736 support@wiredfaculty.com

Sponsor Area

Continuity And Differentiability

Question

Wired Faculty App

Differentiate the following w.r.t.x: $sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 minus 2 straight x squared end fraction close parentheses$

Solution

Let space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 minus 2 straight x squared end fraction close parentheses semicolon space Put space straight x equals sin space straight theta therefore space straight y equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator 1 minus 2 sin squared space straight theta end fraction close parentheses equals sec to the power of negative 1 end exponent open parentheses fraction numerator 1 over denominator cos space 2 space straight theta end fraction close parentheses equals sec to the power of negative 1 end exponent left parenthesis sec space 2 straight theta right parenthesis equals 2 straight theta therefore space straight y equals 2 space sin to the power of negative 1 end exponent straight x space rightwards double arrow space dy over dx equals fraction numerator 2 over denominator square root of 1 minus straight x squared end root end fraction

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation