How do you find $\frac{d y}{d x}$ $(dy)/(dx)$ given $x^{2} sin x + y^{2} cos y = 1$ $x^2sinx+y^2cosy=1$ ?

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Answer 1 · 2016-11-22T09:12:39

$\frac{d y}{d x} = - \frac{x (2 sin x + x cos x)}{y (2 cos y - y sin y)}$ $(dy)/(dx)=-(x(2sinx+xcosx))/(y(2cosy-ysiny))$

differentiate both sides with respect to $x$ $""x""$ . For the terms on the $L H S$ $" "LHS$ the product rule will have to be employed twice.

$\frac{d}{d x} (x^{2} sin x + y^{2} cos y) = \frac{d}{d x} (1)$ $d/(dx)(x^2sinx+y^2cosy)=d/(dx)(1)$

$\frac{d}{d x} (x^{2} sin x) + \frac{d}{d x} (y^{2} cos y) = \frac{d}{d x} (1)$ $d/(dx)(x^2sinx)+d/(dx)(y^2cosy)=d/(dx)(1)$

$2 x sin x + x^{2} cos x + 2 y \frac{d y}{d x} cos y + y^{2} (- sin y) \frac{d y}{d x} = 0$ $2xsinx+x^2cosx+2y(dy)/(dx)cosy+y^2(-siny)(dy)/(dx)=0$

now rearrange for $\frac{d y}{d x}$ $""(dy)/(dx)$

$\frac{d y}{d x} (2 y cos y - y^{2} sin y) = - (2 x sin x + x^{2} cos x)$ $(dy)/(dx)(2ycosy-y^2siny)=-(2xsinx+x^2cosx)$

$\frac{d y}{d x} = - \frac{x (2 sin x + x cos x)}{y (2 cos y - y sin y)}$ $(dy)/(dx)=-(x(2sinx+xcosx))/(y(2cosy-ysiny))$

How do you find (dy)/(dx)dydx(dy)/(dx) given x^2sinx+y^2cosy=1x2sinx+y2cosy=1x^2sinx+y^2cosy=1?