Let f(x,y)=(x^2)sin(xy).What is the directional derivative of f(x,y) at (1,pi) along direction u=(3i-4j)/5

Thanks for helping~

The answer will be 3/5 times the partial derivative of f with respect to x, minus 4/5 times the partial derivative of f with respect to y, evaluated at the point x=1, y=pi.

OK. So, the partial derivative of f w/r/t x is
(2x)sin(xy) + (x^2)y*cos(xy). At (1,pi), this is 0 + pi*(-1).
The partial derivative of f w/r/t y is
(x^3)cos(xy). At (1,pi), this is (1)*(-1).
So the directional derivative along the direction given by the unit vector is
- (3pi + 4)/5.

The unit vector on that direction is = u/ IuI
IuI = (1/5) sqrt (9+16) = 5/5 = 1
= 3i-4j
DFd = Nabla F dot ( Directional derivative)

Nabla F= DF/dx i +DF/dy j
Nabla F= (2x sin(xy) +x^2 y cos (xy) ) i + x^3 cosxy j
At (1, pi)

Nabla F=( 0+ pi (-1) ) i + (-1) j
Nabla F= -pi i - j
Nabla F dot = ( -pi i- j) dot (3i-4j) = -3pi +4