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Formation of a Differential Equation from a General Solution

If x=tane-y...

Question

If $x = t a n (e^{- y})$ , then show that $\frac{d y}{d x} = \frac{e^{- y}}{1 + x^{2}}$ .

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Solution

$x = tan (e^{- y})$
${tan}^{- 1} x = e^{- y}$
On differentiating, we get
$\frac{1}{1 + x^{2}} = - e^{- y} \times \frac{d y}{d x}$
So, $\frac{d y}{d x} = \frac{- e^{y}}{1 + x^{2}}$

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