Final answer:
To evaluate (dy)/(dx) for the given equation at the point (0, (3π)/(2)), differentiate both sides of the equation with respect to x using the chain rule. Substitute the values x=0 and y=(3π)/(2) into the resulting expression to find the value of (dy)/(dx). The value of (dy)/(dx) at the given point is 0.
Step-by-step explanation:
To evaluate (dy)/(dx) for the equation cos(y+x+x) - y sin(y+x+x²) (y'+1+2x) = 3x² at the point (0, (3π)/(2)), we first differentiate both sides of the equation with respect to x using the chain rule. This will give us an expression involving (dy)/(dx). Next, we substitute the values x=0 and y=(3π)/(2) into the resulting expression to find the value of (dy)/(dx) at the given point.
Let's differentiate both sides of the equation:
Step 1: Differentiate cos(y+x+x) - y sin(y+x+x²) (y'+1+2x) = 3x²
Using the chain rule, we have:
-sin(y+x+x) (1+y'+1+2x)-ysin(y+x+x²) (1+2y'+2x)-ysin(y+x+x²) (y''+2)-ysin(y+x+x²)(y'+1+2x) = 6x
Substituting x=0 and y=(3π)/(2):
Simplifying the equation, we get:
(dy)/(dx) = 6x / (-sin(y+x+x) (1+y'+1+2x)-ysin(y+x+x²) (1+2y'+2x)-ysin(y+x+x²) (y''+2)-ysin(y+x+x²)(y'+1+2x))
Substituting x=0 and y=(3π)/(2), we get:
(dy)/(dx) = 0 / (-sin((3π)/(2)+0+0) (1+0+1+0)-((3π)/(2))sin((3π)/(2)+0+0²) (1+2(0)+2(0))-((3π)/(2))sin((3π)/(2)+0+0²) ((3π)/(2))''+2)-((3π)/(2))sin((3π)/(2)+0+0²)((3π)/(2))'+1+2(0)))
Simplifying further, the value of (dy)/(dx) at the point (0, (3π)/(2)) is:
(dy)/(dx) = 0 / (-sin((3π)/(2)) (1+1)-((3π)/(2))sin((3π)/(2)))-((3π)/(2))sin((3π)/(2)) (0+2)-((3π)/(2))sin((3π)/(2)) ((3π)/(2))''+2)-((3π)/(2))sin((3π)/(2))((3π)/(2))'+1))