Let there be a function $z=x^3+xy+y^2$. Is there a vector $\vec U$ that in the point $(1,1)$ the function derivative in the vector direction is equal to 6?
Now I know that the $|\nabla F(1,1)|$ is $\sqrt{4^2+5^2} = \sqrt{41} = 6.4 > 6$. So there is a vector $\vec U$ that equal to 6.
How can I find it?
Thank!