What is directional derivative of the function f(x,y)=x^2y^3-4y at the point (1,2) in the direction of the vector 2i+3j ?

We'll note the given vector as v = 2i +
3j.

For the beginning, we'll have to determine the gradient
vector at the point (1 , 2).

Since the function is of 2
variables, the gradient of the function is the vector
function:

Grad f(x,y) = [df(x,y)/dx]*i +
[df(x,y)/dy]*j

[df(x,y)/dx] and [df(x,y)/dy] are the
partial derivatives of the function.

Grad f(x,y) = 2x*y^3*i
+ (3x^2*y^2 - 4)*j

Grad f(1,2) = 2*2^3*i + (3*2^2 -
4)*j

Grad f(1,2) = 16*i +
8*j

Now, we'll have to determine the unit vector in the
direction of v:

u = v/|v|

|v|
= sqrt(2^2 + 3^2)

|v| = sqrt(4 +
9)

|v| = sqrt 13

u =
2*i/sqrt13 + 3*j/sqrt13

Now, we have all elements to
calculate the directional derivative at the point
(1,2):

Df(1,2) = Grad f*u = (16*i + 8*j)*(2*i/sqrt13 +
3*j/sqrt13)

Since i*i = i^2 = 1 and i*j = 0, we'll
get:

Df(1,2) = 32/sqrt13 +
24/sqrt13

Df(1,2) =
56/sqrt13

Df(1,2) =
56*sqrt13/13

The directional derivative of
the given function, at the point (1,2), is : Df(1,2) =
56*sqrt13/13.