Given f(x,y)=x^3+x^2*y^3-2y^2, find fx(2,1) and fy(2,1)?

Expert Answers
hala718 eNotes educator| Certified Educator

f(x,y)= x^3+x^2*y^3-2y^2

find fx(2,1) and fy(2,1)

First we need to calculate fx(x,y) and fy(x,y)

For fx(x,y) we differentiate with respect to x.

==> fx(x,y) = 3x^2 +(2y^3)x

Now substitute with (2,1)

==> fx(2,1)= 3(4) +2(1)2 = 12+4= 16

Now, for fy(x,y), we differentiate with respect to y

==> fy(x,y) = 3(x^2)y^2 -4y

Substitute with (2,1)

==> fy(2,1) = 3(4)(1) -4(1) = 12-4= 8

neela | Student

f(x,y) = x^3+x^2*y^3-2y^2.

To find fx(2,1) and fy(2,1).

Solution:

fx(x,y) is the partial derivative of f(x,y). Here we take the partial derivative of f(x,y) with respct to x treating x as variable and y as constant.

So fx(x,y) = d/dx{x^3 +x^2*y^3-2y^2}

=d/dx(x^3)+y^3*(d/dx)x^2- (d/dx)(2y^2)

=3x^2+y^3*2x -0 = 3x^2+2x*y^3

So, fx(x,y) = 3x^2+2xy^3

Therefore, fx(2,1) = 3*2^2 +2*2*1^3 = 12 +4 = 16

To find fy(x,y) : Here we take the derivative of f(x,y) with respect to y treating y as varible and x as constant.So,

fy(x,y) = d/dy{x^3+x^2*y^3-2y^2}

= d/dy(x^3) + x^2*(d/dy)y^3- (d/dy)(2y^2)

=0 + x^2*3y^3-4y

fy(x,y) = 3x^2*y^2-4y.

fy(2,1) = 3*2^2*1^2-4*1 = 12-4 = 8

 

giorgiana1976 | Student

To calculate the values of the partial derivatives, we have to find first their expressions.

First, we'll differentiate with respect to x, holding y constant and we'll get:

fx(x,y)=3x^2+2x*y^3

Now, we'll differentiate with respect to y, keeping x constant.

fy(x,y)=3x^2*y^2-4y

Now, we'll calculate fx(2,1):

fx(2,1) = 3*2^2+2*2*1^3 = 16

We'll calculate fy(2,1):

fy(2,1)=3*2^2*1^2-4*1 = 8