IIT JEE , AIEEE Forum - Mathematics , Differential Calculus - Differentiation question--just solve it

nknikhilesh1

If x^y=e^x-y , prove that dy/dx=logx / (1+logx)²

chinmay_saxena01

take log on both sides so u will get
ylogx = x-yloge
ylogx+y=x
y= x/(1+logx)
now differentiate w.r.t. x u will get....
dy/dx = logx/(1+logx)^2

ganesha1991

taking los to the base e both the sides

logx^y =log[ e^(x-y)]
y logx = x-y
differentiatintg both sides

dy/dxlogx+ Y/x = 1 - dy/dx
dy.dx ( logx+1) = 1-y/x
= [ 1-y/x] / logx+1
but ylogx =x-y
y ( logx+1) = x
y = x / (logx+1)

so dy/dx = [ 1- x / (logx+1)x ] / (logx+1)
= logx +1 -1 / (logx+1)^2
= logx / (logx+1)^2

chinmay_saxena01

here it is,,,,,(1+logx)d/dxx-xd/dx(1+logx)/(1+logx)^2
now it is dy/dx = 1+logx-1/(1+logx(^2
if i helped u plzz rate me,,,,

bhavsar

x^y=e^x-y

so that,

e^ylogx=e^x-y

^therefore,

^{ylogx = x-y.........................................}(1)

logx+1=x/y............................(2)

take d/dx at (1)

sothat,

y/x + logxdy/dx=1-dy/dx

dy/dx=x-y/x(logx +1)

from (1)&(2)

dy/dx=logx/(1+logx)²

A_AMBASTHA

HE IS RIGHT....FOLLOW THIS WAY...

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