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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)
if y to the power x = x to the power y, prove that dy/dx=y/x(y-xlogy/x-ylogx)
take logarihtm on both sides and differenciate
could u pls defferentiate it for me... i tried but couldn''t get the 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)
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)
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)
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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