Question

Derive

Sinxy = x² + 2xy​

Answer 1

`(dy)/(dx) = (2x +2 y- ycosxy)/(x(cosxy-2))`

Explanation:

Step(i):-

Given

sin xy = x² + 2xy ...(i)

Apply

`(d)/(dx) UV = U^(l) V + U V^(l)`

`(d)/(dx) x^(n) = nx^(n-1)`

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

`cosxy (d)/(dx) (xy) = 2x + 2( x(dy)/(dx) + y(1))`

`cosxy (x(dy)/(dx) +y(1)) = 2x + 2( x(dy)/(dx) + y(1))`

`cosxy (x(dy)/(dx)) + ycosxy = 2x + 2( x(dy)/(dx)) +2 y)`

`cosxy (x(dy)/(dx)) - 2( x(dy)/(dx)) = 2x +2 y- ycosxy`

`(cosxy-2) (x(dy)/(dx)) = 2x +2 y- ycosxy`

`(dy)/(dx) = (2x +2 y- ycosxy)/(x(cosxy-2))`