Question

Given the function f(x,y)= arctan(xy) (A) Find the gradient vector f(x,y) (B) Compute the directional derivative of f at the point (1,2) in the direction of the vector

v = 5i + 10j

Answer 1

$\displaystyle f(x,y) = arc\tan(xy)\\ \frac{\partial f(x,y)}{\partial x}=\frac{y}{1+(xy)^2}, \qquad \frac{\partial f(x,y)}{\partial x}=\frac{2}{5}\\ \frac{\partial f(x,y)}{\partial y}=\frac{x}{1+(xy)^2}, \qquad \frac{\partial f(x,y)}{\partial x}=\frac{1}{5}\\ \text{Next, we find the unit vector of the given direction}\\ u=\frac{5i+10j}{\sqrt{125}}= \frac{5i+10j}{5\sqrt{5}}\\ \text{the directional derivative is given by $\nabla f(x)\cdot u$}\\ = \frac{3\sqrt{5}}{25}$