If y = xcot x + 2x2 – 3/x2 + x + 2, find dy/dx

Mention each and every step

Asked by haroonrashidgkp | 19th Jun, 2018, 10:57: PM

Expert Answer:

y = xcot x + (2x2 – 3)/(x2 + x + 2)
begin mathsize 16px style straight y equals straight x to the power of cotx plus fraction numerator 2 straight x squared minus 3 over denominator straight x squared plus straight x plus 2 end fraction
straight y equals straight u plus straight v
straight u equals straight x to the power of cotx space space space and space straight v equals fraction numerator 2 straight x squared minus 3 over denominator straight x squared plus straight x plus 2 end fraction
logu equals cotxlogx
1 over straight u du over dx equals cotx cross times 1 over straight x plus logx cross times open parentheses negative cosec squared straight x close parentheses
du over dx equals straight x to the power of cotx space open parentheses cotx cross times 1 over straight x minus logx cross times cosec squared straight x close parentheses
straight v equals fraction numerator 2 straight x squared minus 3 over denominator straight x squared plus straight x plus 2 end fraction
dv over dx equals fraction numerator open parentheses straight x squared plus straight x plus 2 close parentheses cross times 4 straight x minus open parentheses 2 straight x squared minus 3 close parentheses open parentheses 2 straight x plus 1 close parentheses over denominator open parentheses straight x squared plus straight x plus 2 close parentheses squared end fraction
dv over dx equals fraction numerator 4 straight x cubed plus 4 straight x squared plus 8 straight x minus open parentheses 4 straight x cubed plus 2 straight x squared minus 6 straight x minus 3 close parentheses over denominator open parentheses straight x squared plus straight x plus 2 close parentheses squared end fraction
dv over dx equals fraction numerator 4 straight x squared minus 2 straight x squared plus 8 straight x plus 6 straight x plus 3 over denominator open parentheses straight x squared plus straight x plus 2 close parentheses squared end fraction
dv over dx equals fraction numerator 2 straight x squared plus 14 straight x plus 3 over denominator open parentheses straight x squared plus straight x plus 2 close parentheses squared end fraction
dy over dx equals straight x to the power of cotx space open parentheses cotx cross times 1 over straight x minus logx cross times cosec squared straight x close parentheses plus fraction numerator 2 straight x squared plus 14 straight x plus 3 over denominator open parentheses straight x squared plus straight x plus 2 close parentheses squared end fraction end style

Answered by Sneha shidid | 20th Jun, 2018, 10:02: AM