find dy/dx

y = (log x)x + (x)log x

mention each and every formula and minute details

Asked by haroonrashidgkp | 2nd Jun, 2018, 08:30: PM

Expert Answer:

y = (logx)x + xlogx
y = u + v
u = (logx)x 
Taking log on both sides,
log u = xlog(logx)
begin mathsize 16px style 1 over straight u du over dx equals straight x cross times fraction numerator 1 over denominator log begin display style straight x end style end fraction cross times 1 over straight x plus log left parenthesis logx right parenthesis
fraction numerator begin display style du end style over denominator begin display style dx end style end fraction equals left parenthesis logx right parenthesis to the power of straight x open parentheses fraction numerator 1 over denominator log begin display style straight x end style end fraction plus log left parenthesis logx right parenthesis close parentheses end style
 
v = xlogx
Taking log on both sides,
logv = logx×logx
logv = (logx)2 
begin mathsize 16px style 1 over straight v dv over dx equals 2 logx cross times 1 over straight x
fraction numerator begin display style dv end style over denominator begin display style dx end style end fraction equals straight x to the power of logx cross times fraction numerator 2 logx over denominator straight x end fraction end style
 y = u + v
begin mathsize 16px style dy over dx equals du over dx plus dv over dx end style
Error converting from MathML to accessible text.

Answered by Sneha shidid | 4th Jun, 2018, 10:20: AM