dy/dx = +

plz correct me if I'm wrong

y = (x + 1/x) + (x)^1+1/x





solution-



taking log on both sides



logy = log(x+1/x)^x +log(x)^1+1/x



differentiating w.r.t to x



1/y*dy/dx = xlog(x+1/x)+1+1/x logx



dy/dx= y* x log(x+1/x )+(1+1/x)logx



now substitute the value of y

dy/dx = (x+1/x)+(x)^(1+1/x)* x log(x+1/x)+(1+1/x)logx

See akku,I think u didn't understand what I nudged you.

y=[1+1/x]^x+(x)^1+1/x

Taking log on both sides will give you,

lny=ln[(1+1/x)^x+(x)^1+1/x] and not ln(1+1/x)^x+ln(x)^1+1/x.

Hope u got it.