Explanation
Let y=y1+y2 ⇒dydx=dy1dx+dy2dx Now y1=(1+1x)x⇒logy1=xlog(1+1x)
Differentiating both side w.r.t x
⇒1y1dy1dx=log(1+1x)+x.11+1x.(−1x2)
⇒dy1dx=y1[log(1+1x)−11+x] And y2=x1+1x⇒logy2=(1+1x)logx
Differentiating both sides,
⇒1y2dy2dx=(1+1x)1x+(logx)(−1x2)
⇒dy2dx=y2[(1+x)−logxx2] ∴dydx=(1+1x)[log(1+1x)−11+x]+x1+1x[(1+x)−logxx2]
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