Question

Differentiate the following functions w.r.t.x: $tan to the power of negative 1 end exponent open parentheses fraction numerator cos space straight x over denominator 1 plus sin space straight x end fraction close parentheses$

Name: Wired Faculty - The Best Learning App
Rating: 4.6 (1864 reviews)

Solution

Let space space straight y equals tan to the power of negative 1 end exponent open parentheses fraction numerator cos space straight x over denominator 1 plus sin space straight x end fraction close parentheses equals tan to the power of negative 1 end exponent open square brackets fraction numerator cos squared begin display style straight x over 2 end style minus sin squared begin display style straight x over 2 end style over denominator cos 2 begin display style straight x over 2 end style plus sin squared begin display style straight x over 2 end style plus 2 sin begin display style straight x over 2 end style cos begin display style straight x over 2 end style end fraction close square brackets space space space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator open parentheses cos begin display style straight x over 2 end style plus sin begin display style straight x over 2 end style close parentheses open parentheses cos begin display style straight x over 2 end style minus sin begin display style straight x over 2 end style close parentheses over denominator open parentheses cos begin display style straight x over 2 end style plus sin begin display style straight x over 2 end style close parentheses squared end fraction close square brackets space space space space space space space space space space space space equals tan to the power of negative 1 end exponent open square brackets fraction numerator cos begin display style straight x over 2 end style minus sin begin display style straight x over 2 end style over denominator cos begin display style straight x over 2 end style plus sin begin display style straight x over 2 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets fraction numerator 1 minus tan begin display style straight x over 2 end style over denominator 1 plus tan begin display style straight x over 2 end style end fraction close square brackets equals tan to the power of negative 1 end exponent open square brackets tan open parentheses straight pi over 2 minus straight x over 2 close parentheses close square brackets equals straight pi over 2 minus straight x over 2 therefore space dy over dx equals negative 1 half

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Differentiate the following functions w.r.t.x: $tan to the power of negative 1 end exponent open parentheses fraction numerator cos space straight x over denominator 1 plus sin space straight x end fraction close parentheses$

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location