Question

Differentiate the following functions by substitutions method : $tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 1 plus square root of 1 plus straight x squared end root end fraction close parentheses$

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

For Daily Free Study Material Join wiredfaculty

Continuity And Differentiability

Differentiate the following functions by substitutions method : $tan to the power of negative 1 end exponent open parentheses fraction numerator straight x over denominator 1 plus square root of 1 plus straight x squared end root end fraction close parentheses$

Some More Questions From Continuity and Differentiability Chapter

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Phone Number

Email Address

Location