The correct option is A 2(1−ln2)
We have,
yx−xy=1 ⋯(1)
At x=1,
y=1+1=2
Now, from (1)
exlny−eylnx=1
On differentiating w.r.t. x, we get
exlny(xy⋅dydx+lny)−eylnx(dydxlnx+yx)=0
⇒yx(xy⋅dydx+lny)−xy(dydxlnx+yx)=0
On putting x=1 and y=2, we get
2(12⋅dydx+ln2)−(0+2)=0
⇒dydx=2−2ln2
=2(1−ln2)