If x-ex-y- prove that dydx-x-yx log x

Applying Log on both sides log x =xy log e y log x = x Differentiating wrt x log x dydx + yx = 1 #160; log x dydx = x yx dydx = x ...

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Inverse of a Function

If x=ex/y, ...

Question

If $x = e^{\frac{x}{y}}$ , prove that $\frac{d y}{d x} = \frac{x - y}{x log x}$

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Solution

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x = e^{x / y}

\frac{d y}{d x} = \frac{x - y}{x log x}

x^{y} = e^{x - y}

\frac{d y}{d x} = \frac{y log x}{x (log x + 1)}

e^{x} + e^{y} = e^{x + y}

\frac{d y}{d x} + e^{y - x} = 0

y = e^{x} + e^{- x}

\frac{d y}{d x} = \sqrt{y^{2} - 4}

e^{x - y} = x^{y}

\frac{d y}{d x} = \frac{log x}{{[log e x]}^{2}}

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