Question

Derive Sinxy = x² + 2xy​

Answer 1

Answer:

    $\frac{dy}{dx} = \frac{2x +2 y- ycosxy}{x(cosxy-2)}$

Step-by-step explanation:

Step(i):-

Given

        sin xy = x² + 2xy ...(i)

Apply

       $\frac{d}{dx} UV = U^{l} V + U V^{l}$

     $\frac{d}{dx} x^{n} = nx^{n-1}$

Differentiating equation (i) with respective to 'x', we get

      $cosxy \frac{d}{dx} (xy) = 2x + 2( x\frac{dy}{dx} + y(1))$

      $cosxy (x\frac{dy}{dx} +y(1)) = 2x + 2( x\frac{dy}{dx} + y(1))$

     $cosxy (x\frac{dy}{dx}) + ycosxy = 2x + 2( x\frac{dy}{dx}) +2 y)$

   $cosxy (x\frac{dy}{dx}) - 2( x\frac{dy}{dx}) = 2x +2 y- ycosxy$

   $(cosxy-2) (x\frac{dy}{dx}) = 2x +2 y- ycosxy$

       $\frac{dy}{dx} = \frac{2x +2 y- ycosxy}{x(cosxy-2)}$