If sin(x+y)=3x−2y, then dydx=
To find dy/dx given sin(x + y) = 3x – 2y, we will differentiate both sides of the equation with respect to x using the chain rule
To find dy/dx given sin(x + y) = 3x – 2y, we will differentiate both sides of the equation with respect to x using the chain rule.
First, let’s rewrite the equation to make it easier to differentiate:
sin(x + y) = 3x – 2y
Using the chain rule, we differentiate both sides of the equation with respect to x:
d/dx(sin(x + y)) = d/dx(3x – 2y)
To differentiate sin(x + y), we use the chain rule:
cos(x + y) * (d/dx(x + y)) = 3 – 0
Differentiating x with respect to x gives us 1, and differentiating y with respect to x gives us dy/dx.
cos(x + y) * (1 + dy/dx) = 3
Now, let’s isolate dy/dx by rearranging the equation:
cos(x + y) + cos(x + y) * dy/dx = 3
cos(x + y) * dy/dx = 3 – cos(x + y)
Finally, divide both sides by cos(x + y) to obtain the value of dy/dx:
dy/dx = (3 – cos(x + y)) / cos(x + y)
Therefore, the value of dy/dx is (3 – cos(x + y)) / cos(x + y).
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