Answer:
[tex]\frac{dy}{dx} = \frac{2x +2 y- ycosxy}{x(cosxy-2)}[/tex]
Step-by-step explanation:
Step(i):-
Given
sin xy = x² + 2xy ...(i)
Apply
[tex]\frac{d}{dx} UV = U^{l} V + U V^{l}[/tex]
[tex]\frac{d}{dx} x^{n} = nx^{n-1}[/tex]
Differentiating equation (i) with respective to 'x', we get
[tex]cosxy \frac{d}{dx} (xy) = 2x + 2( x\frac{dy}{dx} + y(1))[/tex]
[tex]cosxy (x\frac{dy}{dx} +y(1)) = 2x + 2( x\frac{dy}{dx} + y(1))[/tex]
[tex]cosxy (x\frac{dy}{dx}) + ycosxy = 2x + 2( x\frac{dy}{dx}) +2 y)[/tex]
[tex]cosxy (x\frac{dy}{dx}) - 2( x\frac{dy}{dx}) = 2x +2 y- ycosxy[/tex]
[tex](cosxy-2) (x\frac{dy}{dx}) = 2x +2 y- ycosxy[/tex]
[tex]\frac{dy}{dx} = \frac{2x +2 y- ycosxy}{x(cosxy-2)}[/tex]
