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प्रश्न
Find the derivative of `y = log x + 1/x` with respect to x.
उत्तर
`y = log x + 1/x`
On differentiating both sides, w.r.t. x
`dy/dx = (d(logx))/dx + d/dx(1/x)`
= `1/x + ((-1)/x^2)`
= `1/x - 1/x^2`
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