Mathematics

Mathematics

Problem

If yln⁡y=xy\ln y = x, then dydx\dfrac{dy}{dx} equals a) 1y(ln⁡y+1)\displaystyle\frac{1}{y(\ln y + 1)} b) 1yln⁡y+1\displaystyle\frac{1}{y\ln y + 1} c) 1ln⁡y+1\displaystyle\frac{1}{\ln y + 1} d) 1ln⁡y\displaystyle\frac{1}{\ln y}

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