Test:43 QNo:10

Q10 find dy/dx at x=e, y=e^2x/logx

a) (e-1)e^2e/e
b) o
c) 1
d) (e²-1)e^2e/e

Correct Answer: (a)

plz someone show the working

Given y log x = e^2x

so y . [ d/dx { log x } ] + log x . [ d/dx { y } ] = d/dx { e^2x }

or , y / x + dy /dx { log x} = 2 e^2x

or , after simplification , dy / dx = 2 e^2x - { y / x }log x

So dy / dx at x = e ,

is 2 e^2e - e^{2e - 1}1 or , ( 2e - 1 ) e^2ee

I really think the answer's wrong :( :( :(

yeah even i m getting this ans