(1 point) Calculate the directional derivative of f(x, y) = cot-'(xy) in the direction of v = -4i + 5j at the point P = (0.5, 0.5).
Remember to normalize the direction vector.
DS(0.5, 0.5) =

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

$655.61 which is option c

step-by-step explanation:

price of house =   $146,000

initial rate = 3.5%

mortgaged paid in this = 3.5 % of $146,000

                                        =

                                        =$5110

the total years would be = 7

total mortgaged paid = 7 * $5110

                                      =$35770

remaining money = $146000-35770

                                =$110230

new mortgaged payment = 5 % of 110230

                                            =  

                                            =$655.61