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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