Find y’ for y=y(x) defined implicitly for 2y^2-x^3-7=0
To find y', we need to differentiate both sides of the equation with respect to x.
Differentiating the left side:
d/dx (2y^2) = d/dx (x^3) + d/dx (7)
Using the chain rule, we have:
2 * d/dx (y^2) = 3x^2 + 0
Differentiating y^2 with respect to y, we have:
2y * dy/dx = 3x^2
Now, solving for dy/dx, we get:
dy/dx = (3x^2) / (2y)
Therefore, the derivative of y with respect to x, y', is given by:
y' = (3x^2) / (2y)