Question

y³-xy²+cos x y =2
determinar la derivada de la función

Answer 1

Explicación paso a paso:

$\frac{dy}{dx} [y³-xy²+ \cos(xy) ] = \frac{dy}{dx} [2]$

$\frac{dy}{dx} [y³ ] - \frac{dy}{dx} [xy²] + \frac{dy}{dx} [ \cos(xy) ] = 0$

$3 {y}^{2} \: y^{'} - [(1)y² + x(2y\: y^{'})] - [ \sin(xy) ] (y + x\: y^{'}) = 0$

$3 {y}^{2} \: y^{'} - y² - 2xy\: y^{'} - y\sin(xy) - x\ \sin(xy) \: y^{'}) = 0$

$3 {y}^{2} \: y^{'} - 2xy\: y^{'} - x\ \sin(xy) \: y^{'}) = {y}^{2} + y \sin(xy)$

$y^{'}(3 {y}^{2} - 2xy - x\ \sin(xy) )= {y}^{2} + y \sin(xy)$

$y^{'}= \frac{ {y}^{2} + y \sin(xy) }{3 {y}^{2} - 2xy - x\ \sin(xy)}$

Por tanto:

$\frac{dy}{dx} = \frac{ {y}^{2} + y \sin(xy) }{3 {y}^{2} - 2xy - x\ \sin(xy)}$