PROBLEMS ON PARTIAL DERIVATES IN THREE VARAIABLES

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 :

If

U(x, y, z) = (x2+y2)/xy + 3z2y

find (∂u/∂x), (∂u/∂y) and (∂u/∂z)

Solution :

U(x, y, z) = (x2+y2)/xy + 3z2y

Using quotient rule, we get

Differentiate with respect to x, so keep y and z as constants. 

∂u/∂x = [xy(2x)-(x2+y2)y]/(xy)2 + 0

∂u/∂x = [2x2y-x2y+y3]/(xy)2

∂u/∂x = (x2y+y3)/(xy)2

∂u/∂x = (x2+y2)/(x2y)

Differentiate with respect to y, so keep x and z as constants. 

∂u/∂y = [xy(2y)-(x2+y2)x]/(xy)2+3z2(1)

∂u/∂y = [2xy2-x3-xy2]/(xy)2+3z2

∂u/∂y = (xy2-x3)/(xy)2+3z2

∂u/∂y = (y2-x2)/(xy2)+3z2

Differentiate with respect to z, so keep x and y as constants. 

∂u/∂z = 0+3(2z)y

∂u/∂z = 6zy

Problem 2 :

If U(x, y, z) = log (x3+y3+z3), find 

(∂u/∂x) + (∂u/∂y) + (∂u/∂z)

Solution :

U(x, y, z) = log (x3+y3+z3)

Finding ∂u/∂x :

Differentiate with respect to x. Treat y and z as constant.

∂u/∂x = [1/(x3+y3+z3)] (3x2)

∂u/∂x = [3x2/(x3+y3+z3)]  ------(1)

∂u/∂y = [1/(x3+y3+z3)] (3y2)

∂u/∂y = [3y2/(x3+y3+z3)]  ------(2)

∂u/∂z = [1/(x3+y3+z3)] (3z2)

∂u/∂y = [3z2/(x3+y3+z3)]  ------(3)

(1) + (2) + (3)

(∂u/∂x) + (∂u/∂y) + (∂u/∂z) 

= [3x2/(x3+y3+z3)] + [3y2/(x3+y3+z3)] + [3z2/(x3+y3+z3)]

=  3(x2+y2+z2)/(x3+y3+z3)

Problem 3 :

For each of the following functions find gxy, gxx, gyy,and gyx.

(i)  g(x, y) = xey+3x2y

Solution :

Finding gx :

Differentiating respect to x.

gx = ey+3(2x)y

gx = ey+6xy

Finding gy :

Differentiating respect to y.

gy = xey+3x2(1)

gy = xey+3x2

(i) Finding gxy :

gxy = ey+6x(1)

gxy = ey+6x

(ii) Finding gxx :

Differentiating respect to x.

gx = ey+6xy

gxx = 0+6(1)y

gxx = 6y

(iii) Finding gyy :

gy = xey+3x2

gyy = xey+0

gyy = xey

(iv) Finding gyx :

gy = xey+3x2

gyx = (1)ey+3(2x)

gyx = ey + 6x

(ii)  g(x, y) = log(5x+3y)

Solution :

Finding gx :

gx  =  [1/(5x+3y)] (5)

gx  =  5/(5x+3y)

Finding gy :

gy  =  [1/(5x+3y)] (3)

gy  =  3/(5x+3y)

(i) Finding gxy :

From gx  =  5/(5x+3y)

Differentiate with respect to y.

gxy = [-5/(5x+3y)2](3)

gxy = [-15/(5x+3y)2]

(ii) Finding gxx :

From gx  =  5/(5x+3y)

Differentiating respect to x.

gxx =  [-5/(5x+3y)2](5)

gxx = [-25/(5x+3y)2]

(iii) Finding gyy :

From g= 3/(5x+3y)

Differentiating respect to y.

gyy = -9/(5x+3y)2

(iv) Finding gyx :

From gy  =  3/(5x+3y)

Differentiating respect to x.

gyx = -15/(5x+3y)2

(iii) g(x, y) = x2+3xy-7y+cos(5x)

Solution :

Finding gx :

gx = 2x+3y-0-sinx(5x)(5)

gx = 2x+3y-5sinx(5x)

Finding gy :

gy = 0+3x(1)-7+0

gy = 3x-7

(i) Finding gxy :

From gx = 2x+3y-5sinx(5x)

Differentiate with respect to y.

gxy = 0+3+0

gxy = 3

(ii) Finding gxx :

From gx = 2x+3y-5sinx(5x)

Differentiate with respect to x.

gxx = 2+0-5cos(5x)(5)

gxx = 2-25cos(5x)

(iii) Finding gyy :

From gy = 3x-7

Differentiate with respect to y.

gyy = 0

(iv) Finding gyx :

From gy = 3x-7

Differentiate with respect to x.

gyy = 3

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