How do I find the gradient of (x^2 + y^2 - 1) ^3 =x^2y^3?

Answer 1

Greetings! Lets solve this shall we ?

So, we must find the gradient of this function as it is. Since we do not have a point, we can still find the gradient as a vector. Then, the gradient yields

∇ F(x ,y) = < F_x, F_y>, where F(x, y) = (x²+y²-1)³ - x²y³

We must take the first partial derivatives with respect to x and y such that

∂F/∂x = 3(x²+y²-1)²(2x) - 2xy³ = 6x(x²+y²-1)² - 2xy³

∂F/∂y = 3(x²+y²-1)²(2y) - 3x²y² = 6y(x²+y²-1)² - 3x²y²

Thus, the gradient vector is

∇ F(x, y) = < (6x(x²+y²-1)² - 2xy³), (6y(x²+y²-1)² - 3x²y²)>

I hope this helped!