y^'= 1/(e^y-x)

解答

y^{^{'}} = \frac{1}{e ^{y} - x}

y = ln (x + x^{2} + c_{1}), y = ln (x - x^{2} + c_{1})

求解步骤

y^{'} (x) = \frac{1}{e ^{y} - x}

求解恰当方程:

y = ln (x + x^{2} + c_{1}), y = ln (x - x^{2} + c_{1})

y = ln (x + x^{2} + c_{1}), y = ln (x - x^{2} + c_{1})

\frac{d}{d x} (2 x + 2 y)

\frac{d}{d x} (4 arccos (- 5 x))

\frac{d}{d x} ((sin (x))^{4} + sin (x^{4}))

in v erse l a pl a ce \frac{1}{4 s + 2}

\frac{d}{d θ} (2 θ)