Sponsor Area

Continuity And Differentiability

Question
CBSEENMA12034890

Differentiate the following w.r.t.x :2 square root of cot left parenthesis straight x squared right parenthesis end root

Solution
Let space space space space space straight y equals 2 square root of cot left parenthesis straight x squared right parenthesis end root equals 2 left square bracket cot left parenthesis straight x squared right parenthesis right square bracket to the power of 1 half end exponent
therefore space dy over dx equals 2.1 half left square bracket cot left parenthesis straight x squared right parenthesis right square bracket to the power of negative 1 half end exponent. straight d over dx left parenthesis cot space straight x squared right parenthesis
space space space space space space space space space space space space space equals fraction numerator 1 over denominator square root of cot space straight x squared end root end fraction. left parenthesis negative cosec squared straight x squared right parenthesis. straight d over dx left parenthesis cot space straight x squared right parenthesis
space space space space space space space space space space space space space equals fraction numerator 1 over denominator square root of cot space straight x squared end root end fraction. left parenthesis negative cosec squared straight x squared right parenthesis. left parenthesis 2 straight x right parenthesis equals negative fraction numerator 2 straight x space cosec squared straight x squared over denominator square root of cot space straight x squared end root end fraction

Some More Questions From Continuity and Differentiability Chapter