if y = esinx + (tan x)x , prove that

dy/dx = esin x cos x + (tan x)x [2x cosec 2x + log tan x]

Asked by haroonrashidgkp | 11th Jul, 2018, 12:18: PM

Expert Answer:

y = esinx + (tan x)x 
Let u =  esinx  and v = (tan x)x
u = esinx 
begin mathsize 16px style du over dx equals straight e to the power of sinx cosx end style........(i)
 
v = (tanx)x 
log v = xlog tanx
begin mathsize 16px style 1 over straight v dv over dx equals straight x cross times fraction numerator 1 over denominator tan begin display style straight x end style end fraction cross times sec squared straight x plus log space tanx space
fraction numerator begin display style dv end style over denominator begin display style dx end style end fraction equals open parentheses tanx close parentheses to the power of straight x open parentheses straight x cross times fraction numerator cos begin display style straight x end style over denominator sin begin display style straight x end style end fraction cross times fraction numerator 1 over denominator begin display style cos squared end style begin display style straight x end style end fraction plus log space tanx close parentheses
fraction numerator begin display style dv end style over denominator begin display style dx end style end fraction equals open parentheses tanx close parentheses to the power of straight x open parentheses straight x cross times fraction numerator 1 over denominator sin begin display style straight x end style end fraction cross times fraction numerator 1 over denominator begin display style cos straight x end style end fraction plus log space tanx close parentheses
fraction numerator begin display style dv end style over denominator begin display style dx end style end fraction equals open parentheses tanx close parentheses to the power of straight x open parentheses xcosecxsecx plus log space tanx close parentheses........... left parenthesis ii right parenthesis end style
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Answered by Sneha shidid | 11th Jul, 2018, 02:26: PM