First of all, this can be written as y = xln(1 + 1/x)y=xln(1+1x).
We now use the chain and power rules to differentiate ln(1 + 1/x)ln(1+1x) and the product rule to differentiate the entire function.
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Differentiating ln(1 + 1/x)ln(1+1x):
Let y= lnuy=lnu and u = 1 + 1/xu=1+1x. Then dy/(du) = 1/udydu=1u and (du)/dx = -1/x^2dudx=−1x2.
dy/dx = dy/(du) xx (du)/dxdydx=dydu×dudx
dy/dx= 1/u xx -1/x^2dydx=1u×−1x2
dy/dx = -1/((1 + 1/x)(x^2))dydx=−1(1+1x)(x2)
dy/dx = -1/(x^2 + x)dydx=−1x2+x
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Differentiating y= xln(1 +1/x)y=xln(1+1x):
y' = 1(ln(1 + 1/x)) + (-1/(x^2 + x) xx x)
y' = ln(1 + 1/x) - x/(x(x + 1))
y' = ln(1 + 1/x) - 1/(x + 1)
Hopefully this helps!
