Sum

Find `dy/dx if y = ((logx+1))/x`

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#### Solution

`y=((logx + 1))/x`

Differentiating w.r.t. x, we get

`dy/dx=d/dx[(logx + 1)/x]`

= `(xd/dx(logx + 1) - (logx + 1)d/dx(x))/x^2`

= `(x(1/x + 0) - (logx + 1)(1))/x^2`

= `(1 - logx - 1)/x^2`

=`(-logx)/x^2`

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