Question

How do you find the derivative of `sin(x+y)=xy`?

Answer 1

Start by expanding `sin(x + y)` using the sum and difference identity.

`sin(x + y) = xy`

`sinxcosy + cosxsiny = xy`

By the product rule:

`cosxcosy + sinx xx -siny(dy/dx) -sinxsiny + cosxcosy(dy/dx) = y + x(dy/dx)`

`cosxcosy(dy/dx) - sinxsiny(dy/dx) - x(dy/dx) = sinxsiny - cosxcosy + y`

`dy/dx(cosxcosy - sinxsiny - x)= sinxsiny - cosxcosy + y`

`dy/dx = (sinxsiny - cosxcosy + y)/(cosxcosy - sinxsiny - x)`

By the formula `cosAcosB - sinAsinB = cos(A + B)`, we have:

`dy/dx = (-cos(x+ y) - y)/(cos(x + y) - x)`

`dy/dx= -(cos(x + y) - y)/(cos(x + y)- x)`