FIND THE SECOND PARTIAL DERIVATIVES OF THE FUNCTION

When we find partial derivative of F with respect to x, we treat the y variable as a constant and find derivative with respect to x .

That is, except for the variable with respect to which we find partial derivative, all other variables are treated as constants. That is why we call them as “partial derivative”.

If F has a partial derivative with respect to x at every point of A , then we say that (∂F/∂x) (x, y) exists on A.

Note that in this case (∂F/∂x) (x, y) is again a real-valued function defined on A .

Problem 1 :

Let

w(x, y, z) = 1/√(x2+y2+z2), (x, y, z) ≠ (0, 0, 0).

Show that (2w/∂x2) + (2w/∂y2(2w/∂z2)

Solution :

Given :

w(x, y, z) = 1/√(x2+y2+z2)

∂w/dx  =  -1/2(x2+y2+z2)(3/2)(2x)

∂w/dx  =  -x/(x2+y2+z2)(3/2)

u = -x and v = (x2+y2+z2)(3/2)

u' = -1 and v' = (3/2)(x2+y2+z2)1/2 (2x)

v' = 3x(x2+y2+z2)1/2

(1) + (2) + (3)

=  (2x2-y2-z2-x2+2y2-z2-x2-y2+2z2)/(x2+y2+z2)5/2

=  0

Problem 2 :

If V(x, y) = ex(x cos y - y sin y), then prove that 

2w/∂x2  = ∂2w/∂y2

Solution :

Differentiating with respect to x.

w/∂x  =  ex(cos y - 0) + (x cos y - y sin y)ex

w/∂x =  ex [cos y + x cos y - y sin y]

Differentiate with respect to x.

2w/∂x=  e[0 + x (-sin y) + cos y (1) - 0]

2w/∂x=  e[-x sin y + cos y] ----(1)

Differentiating with respect to y.

w/∂y  =  ex(-xsiny - [y(cosy) + sin y (1)] )

w/∂y =  e[-xsiny - ycosy + sin y] 

Differentiate with respect to x.

∂w/∂y =  e[-xsiny - ycosy + sin y] 

2w/∂y=  e[-xcosy -y(sin y) + cosy(1)+ + cos y

0 + x (-sin y) + cos y (1) - 0]

2w/∂x=  e[-x sin y + cos y] ----(2)

(1)  =  (2)

Problem 3 :

If w(x, y) = xy + sin(xy) , then prove that

2w/∂y∂x  = ∂2w/∂x∂y

Solution :

Differentiating with respect to y. 

∂w/∂y = x(1) + cos(xy)(x) 

∂w/∂y = x + xcos(xy)

∂w/∂y = x[1 + cos(xy)]

Differentiating with respect to x. 

2w/∂y∂x  =  x[0-sin(xy)(y)]+[1 + cos(xy)](1)

2w/∂y∂x  =  -xysin(xy) + 1 + cos(xy)

2w/∂y∂x  =  1 + cos(xy) -xysin(xy) ----(1)

Differentiating with respect to x. 

Given : w(x, y) = xy + sin(xy)

∂w/∂x = y + cos(xy)(y) 

∂w/∂x = y + ycos(xy)

∂w/∂x = y[1+cos(xy)]

Differentiating with respect to y. 

2w/∂x∂y  =  y[0-sin(xy)(x)]+[1 + cos(xy)](1)

2w/∂x∂y  =  -xysin(xy) + 1 + cos(xy) ----(2)

(1)  =  (2)

2w/∂x∂y  =  1 + cos(xy) -xysin(xy)

Problem 4 :

If V(x, y, z)  =  x3+y3+z3+3xyz, show that 

2v/∂y∂z  = ∂2v/∂z∂y

Solution :

Given : V(x, y, z)  =  x3+y3+z3+3xyz

∂v/∂y = 0+3y2+0+3xz

∂v/∂y = 3y2+3xz

2v/∂y∂z = 0+3x(1)

2v/∂y∂z = 3x  ------(1)

Given : V(x, y, z)  =  x3+y3+z3+3xyz

∂v/∂z = 0+0+3z2+0+3xy

∂v/∂z = 3z2+3xy

2v/∂z∂y = 0+3x(1)

2v/∂y∂z = 3x -----(2)

(1)  =  (2)

Problem 5 :

A firm produces two types of calculators each week, x number of type A and y number of type B . The weekly revenue and cost functions (in rupees) are

R(x, y) = 80x + 90y + 0.04xy − 0.05x2 − 0.05y2 and

C(x, y) = 8x + 6y + 2000 respectively.

(i) Find the profit function P(x, y) ,

(ii) Find P/∂x (1200, 1800) and P/∂y (1200, 1800) and interpret these results.

Solution :

To find profit function, we will subtract cost function from revenue function.

P(x, y) = R(x, y) - C(x, y)

= (80x + 90y + 0.04xy − 0.05x2 − 0.05y2)-(8x + 6y + 2000)

P(x, y) = (72x + 84y + 0.04xy − 0.05x2 − 0.05y2- 2000)

P/∂x = 72+0.04y−0.1x

∂P/∂x at (1200, 1800) = 72+0.04(1800)−0.1(1200)

=  72+72-120

=  24

The profit increases.

P/∂y = 84+0.04x-0.1y

∂P/∂x at (1200, 1800) = 84+0.04(1200)-0.1(1800)

=  84+48-180

= 132-144

= -48

The profit decreases.

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