Find $\frac{d y}{d x}$ when $x$ and $y$ are connected by the relation given:
$sin (x y) + \frac{x}{y} = x^{2} - y$

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Solution

$(i)cos(xy)(y+xdfracdydx)+y−xdydxy2=2x−dydx$

$y^{2} c o s (x y) (y + x \frac{d y}{d x}) + y - x \frac{d y}{d x} = 2 x y^{2} - y^{2} \frac{d y}{d x}$

$y^{3} c o s (x y) + x y^{2} \frac{d y}{d x} + y - x \frac{d y}{d x} = 2 x y^{2} - y^{2} \frac{d y}{d x}$

$(x y^{2} - x + y^{2}) \frac{d y}{d x} = 2 x y^{2} - y^{3} c o s (x y) - y$

$\frac{d y}{d x} = \frac{2 x y^{2} - y^{3} c o s (x y) - y}{x y^{2} - x + y^{2}}$

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Similar questions

Find $\frac{d y}{d x}$ when x and y are connected by the relation given.

$s i n (x y) + \frac{x}{y} = x^{2} - y$

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

\frac{d y}{d x}

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

\frac{d y}{d x}

e^{x y} + log (x y) + sin (x y) - 2 = 0

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

\frac{d (x y)}{d x} = 0

\Rightarrow

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

x

y

\frac{d y}{d x}

x = 2 a t^{2}, y = a t^{4}

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