## If sin(x+y)=3x−2y, then dydx=

### To find the derivative dy/dx, we need to differentiate both sides of the equation sin(x+y) = 3x – 2y with respect to x

To find the derivative dy/dx, we need to differentiate both sides of the equation sin(x+y) = 3x – 2y with respect to x.

Applying the chain rule (d/dx[f(g(x))] = f'(g(x)) * g'(x)), we get:

cos(x+y) * (1 + dy/dx) = 3 – 2(dy/dx)

Expanding the equation, we have:

cos(x+y) + cos(x+y) * dy/dx = 3 – 2(dy/dx)

Rearranging terms and isolating dy/dx, we get:

dy/dx + 2(dy/dx) = 3 – cos(x+y) – cos(x+y)

Combining like terms, we have:

3(dy/dx) = 3 – 2cos(x+y)

Now, dividing both sides by 3, we obtain:

dy/dx = (3 – 2cos(x+y))/3

Therefore, dy/dx = (3 – 2cos(x+y))/3 is the derivative we were looking for.

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