+91-8076753736 support@wiredfaculty.com

Sponsor Area

Continuity And Differentiability

Question

Wired Faculty App

Differentiate the following functions w.r.t.x: $cot to the power of negative 1 end exponent square root of fraction numerator 1 plus straight x over denominator 2 end fraction end root$

Solution

Let space space space space space straight y equals cot to the power of negative 1 end exponent square root of fraction numerator 1 plus straight x over denominator 2 end fraction end root therefore space dy over dx equals fraction numerator 1 over denominator 1 plus open parentheses square root of begin display style fraction numerator 1 plus straight x over denominator 2 end fraction end style end root close parentheses squared end fraction straight d over dx open parentheses square root of fraction numerator 1 plus straight x over denominator 2 end fraction end root close parentheses equals negative fraction numerator 1 over denominator 1 plus begin display style fraction numerator 1 plus straight x over denominator 2 end fraction end style end fraction cross times fraction numerator 1 over denominator square root of 2 end fraction straight d over dx left parenthesis square root of 1 plus straight x end root right parenthesis space space space space space space space space space space space space space equals negative fraction numerator 2 over denominator 2 plus 1 plus straight x end fraction cross times fraction numerator 1 over denominator square root of 2 end fraction cross times fraction numerator 1 over denominator 2 square root of 1 plus straight x end root end fraction equals negative fraction numerator 1 over denominator left parenthesis 3 plus straight x right parenthesis square root of 2 square root of 1 plus 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