if y to the power x = x to the power y, prove that dy/dx=y/x(y-xlogy/x-ylogx)

Question

shashank gattani · Answer

take logarihtm on both sides and differenciate

firoz tharayil · Answer

could u pls defferentiate it for me... i tried but couldn''t get the answer..

xyz xz · Answer

taking log on both sides x lny = y lnx diff. both sides let dy/dx = m then lny + (x/y)m = (y/x) + m lnx m= (y/x)/(y-xlogy/x-ylogx)

Yogita Bang · Answer

y^x=x^y xlogy = ylogx (x/y)dy/dx + logy = y/x + logx(dy/dx) dy/dx(x/y-logx) = y/x - logy dy/dx = (y/x - logy)/(x/y-logx) = (y/x){1- (x/y)logy}/{x/y - logx} = y/x(y-xlogy/x-ylogx)

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if y to the power x = x to the power y, prove that dy/dx=y/x(y-xlogy/x-ylogx)

if y to the power x = x to the power y, prove that dy/dx=y/x(y-xlogy/x-ylogx)

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