+91-8076753736 support@wiredfaculty.com

Sponsor Area

Continuity And Differentiability

Question

Wired Faculty App

Differentiate $straight e to the power of log open parentheses straight x plus square root of straight x squared plus straight a squared end root close parentheses end exponent$ w.r.t x.

Solution

Let space straight y equals straight e to the power of log open parentheses straight x plus square root of straight x squared plus straight a squared end root close parentheses end exponent therefore space straight y equals straight x plus square root of straight x squared plus straight a squared end root space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because space straight e to the power of log space straight x end exponent equals straight x right square bracket therefore space dy over dx equals 1 plus fraction numerator 2 straight x over denominator 2 square root of straight x squared plus straight a squared end root end fraction equals 1 plus fraction numerator straight x over denominator square root of straight x squared plus straight a 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