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Question
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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