# 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 ?

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

For the beginning, we'll have to determine the gradient vector at the pointÂ (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.**