# What is the directional derivative of the function at P in the direction of Qf(x, y) = cos(x + y), P(0, pi), Q(pi/2, 0)

*print*Print*list*Cite

You need to find the vector `bar(PQ)` such that:

`bar(PQ) = (x_Q - x_P)(bar i) + (y_Q - y_P)(bar j)`

`bar(PQ) = (pi/2 - 0)(bar i) + (0 - pi)(bar j)`

`bar(PQ) = (pi/2)(bar i) - (pi)(bar j)`

You need to find the unit vector in direction `bar(PQ)` such that:

`bar u = <(pi/2)/(sqrt(pi^2/4 + pi^2)), (-pi)/(sqrt(pi^2/4 + pi^2))>`

`bar u = <(pi/2)/(pisqrt5/2),(-pi)/(pisqrt5/2)>`

`bar u = <sqrt5/5,-2sqrt5/5>`

You need to find the gradient of the function `f(x, y)` such that:

`grad f = <(del f)/(del x),(del f)/(del y)>`

`grad f = <-sin(x+y),-sin(x+y)>`

You need to evaluate the gradient at the point `P(0,pi)` such that:

`grad f(0,pi) = <-sin(pi),-sin(pi)> = <0,0>`

You need to evaluate the directional derivative such that:

`grad f(0,pi)*bar u = <0,0>*<sqrt5/5,-2sqrt5/5> = 0`

**Hence, evaluating the directional derivative of the function, under the given conditions, yields `grad f(0,pi)*bar u = 0.` **