Question Number 10408 by konen last updated on 07/Feb/17

Answered by mrW1 last updated on 08/Feb/17

y^2 =e^(x+y)   ⇒2ln y=x+y  ⇒x=2ln y−y  (dx/dy)=(2/y)−1=((2−y)/y)    (dy/dx)=(1/(dx/dy))=(1/((2−y)/y))=(y/(2−y))    let g=(dy/dx)=(y/(2−y))  (dg/dy)=(1/(2−y))+(y/((2−y)^2 ))=(2/((2−y)^2 ))    (d^2 y/dx^2 )=((d((dy/dx)))/dx)=(dg/dx)=(dg/dy)×(dy/dx)=(2/((2−y)^2 ))×(y/(2−y))=((2y)/((2−y)^3 ))    ⇒answer D