Solution:
Let’s take 
Putting the value of 
So now 
![\frac{d u}{d x}=u\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f2123361806d85afdb3bdcc1d3f6f0de_l3.png "Rendered by QuickLaTeX.com")
![=\left(x+\frac{1}{x}\right)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right] . \ldots \dots \dots](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-b552799c9da2d73b5a46ca4f9082fe3f_l3.png "Rendered by QuickLaTeX.com")
Now again 
![\frac{d v}{d x}=v\left[\frac{1}{x}\left(x+\frac{1}{x}\right)-\frac{1}{x^{2}} \log x\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-5e313b90cbaff79896c74edb1b4eb9b1_l3.png "Rendered by QuickLaTeX.com")
![\frac{d v}{d x}=x^{\left(x+\frac{1}{x}\right)}\left[\frac{1}{x}\left(x+\frac{1}{x}\right)-\frac{1}{x^{2}} \log x\right] \ldots \dots \dots](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-cf52d6a83d6f574a1278b2d5f456ca14_l3.png "Rendered by QuickLaTeX.com")
Putting the values from eq.(2) and eq.(3) in eq.(1),
![\frac{d y}{d x}=\left(x+\frac{1}{x}\right)^{x}\left[\frac{x^{2}-1}{x^{2}+1}+\log \left(x+\frac{1}{x}\right)\right]+x^{\left(x+\frac{1}{x}\right)}\left[\frac{1}{x}\left(x+\frac{1}{x}\right)-\frac{1}{x^{2}} \log x\right]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-a5f59508af50553e6665e3080d9b6eea_l3.png "Rendered by QuickLaTeX.com")
Differentiate the functions with respect to 
Solution:
Let’s take
Putting the value of
So now
Now again
Putting the values from eq.(2) and eq.(3) in eq.(1),
