Looks like f(x, y) = cot⁻¹(xy), as in the inverse cotangent.
If you're like me and you don't remember the derivative of the inverse trig functions, we can derive it as follows. Let y = cot⁻¹(x), so that for appropriate x we can write cot(y) = x. By the chain rule, differentiating both sides gives
-csc²(y) y' = 1
y' = -sin²(y) = -sin²(cot⁻¹(x)) = - 1 / (x² + 1)
Next, recall that the derivative of a function f(x, y) at a point (a, b) in the direction of a vector v is given by
∇ f(a, b) • v
Compute the gradient ∇ f(x, y) :
∇ f(x, y) = - y / ((xy)² + 1) i - x / ((xy)² + 1) j
At the point P, we have
∇ f(x, y) = - 8/17 i - 8/17 j
So, the derivative of f at P along v = - 4 i + 5 j is
(- 8/17 i - 8/17 j) • (- 4 i + 5 j) = - 8/17
