Differentiate:
${x y + y}^{2} = tan x + y$ ?

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Solution

$x y + y ² = t a n x + y$

$\frac{δ x}{δ y} = y - s e c^{2} x$

$\frac{δ x}{δ y} = x + 2 y - 1$

$n o w, \frac{δ x}{δ y} = \frac{- \frac{δ x}{δ y}}{\frac{δ y}{δ x}}$

$= \frac{s e c^{2} x - y}{x + 2 y - 1}$

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Similar questions

\frac{d y}{d x}

x y + y^{2} = tan x + y

x y + y^{2} = tan x + y

\frac{d y}{d x}

\frac{d y}{d x}

tan (x + y) = x^{2} + y^{2}

\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}, y (0) = 0

y^{2} = \exp (x + \frac{x^{2}}{2}) - 1

y^{2} = 1 + C \exp (x + \frac{x^{2}}{2})

y = \tan (x + \frac{x^{2}}{2})

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