Dayv O. answered 05/13/22
Caring Super Enthusiastic Knowledgeable Pre-Calculus Tutor
for dy/dx=y'
note if g(x,y)=x2y, then x and y need to be positive or zero
2y*lnx=ln(2x+8)
2y'*lnx+2y/x=2/(2x+8)=2/x2y
y'=[1/(x2y*lnx)-y/(x*lnx]
y'' is harder
fyi
y''=(1/(ln(x))2)*[(-2x-2y)(1+1/x)+(y/x2)(ln(x)+2)]
unless there is a different answer where y'' is function of x,y