+91-8076753736 support@wiredfaculty.com

Sponsor Area

Continuity And Differentiability

Question

Wired Faculty App

Differentiate the following function w.r.t.x: $cos minus 1 open parentheses fraction numerator straight x over denominator straight x plus 1 end fraction close parentheses$

Solution

Let space space space space space straight y equals cos minus 1 open parentheses fraction numerator straight x over denominator straight x plus 1 end fraction close parentheses therefore space dy over dx equals negative fraction numerator 1 over denominator square root of 1 minus open parentheses begin display style fraction numerator straight x over denominator straight x plus 1 end fraction end style close parentheses squared end root end fraction. straight d over dx open parentheses fraction numerator straight x over denominator straight x plus 1 end fraction close parentheses equals negative fraction numerator straight x plus 1 over denominator square root of left parenthesis straight x plus 1 right parenthesis squared end root end fraction. fraction numerator left parenthesis straight x plus 1 right parenthesis 1 minus straight x.1 over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction space space space space space space space space space space space space space equals negative fraction numerator straight x plus 1 over denominator square root of 1 plus 2 straight x end root end fraction. fraction numerator 1 over denominator left parenthesis straight x plus 1 right parenthesis squared end fraction equals negative fraction numerator 1 over denominator left parenthesis straight x plus 1 right parenthesis square root of 1 plus 2 straight x end root end fraction

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation