PARTIAL DERIVATIVE OF A FUNCTION OF TWO VARIABLES

Find the partial derivatives of the following functions at the indicated points.

Problem 1 :

f (x, y)  =  3x² − 2xy + y² + 5x + 2, (2, −5)

Solution :

f (x, y)  =  3x² − 2xy + y² + 5x + 2

Differentiate with respect to x. Treat y as constant.

∂F/∂x  =  3(2x) - 2(1)y + 5(1) + 0

∂F/∂x  =  6x-2y+5

∂F/∂x at (2, -5)  =  6(2)-2(-5)+5

=  12+10+5

=  27

Differentiate with respect to y. Treat x as constant.

f (x, y)  =  3x² − 2xy + y² + 5x + 2

∂F/∂y  =  0-2x(1)+2y+0+0

∂F/∂y  =  -2x+2y

∂F/∂y at (2, -5)  =  -2(2)+2(-5)

=  -4-10

=  -14

Problem 2 :

g(x, y) = 3x² + y² + 5x + 2, (1, −2)

Solution :

g (x, y)  =  3x² + y² + 5x + 2

Differentiate with respect to x. Treat y as constant.

∂G/∂x  =  3(2x) + 0 + 5(1) + 0

∂G/∂x  =  6x+5

∂G/∂x at (1, -2)  =  6(1)+5

=  11

Differentiate with respect to y. Treat x as constant.

f (x, y)  =  3x² + y² + 5x + 2

∂G/∂y  =  0+2y+0+0

∂G/∂y  =  2y

∂G/∂y at (1, -2)  =  2(-2)

=  -4

Problem 3 :

h(x, y, z) = x sin(xy) + z²x, (2, π/4, 1)

Solution :

h(x, y, z) = x sin(xy) + z²x, (2, π/4, 1)

Differentiate with respect to x. Treat y as constant.

∂H/∂x  =  x(cosy) + sin(xy) (1) + z²(1)

=  2[cos(π/4)] + sin(2⋅(π/4)) + 1²

=  2(1/√2) + sin(π/2) + 1

=  √2 + 1 + 1

=   2 + √2

Differentiate with respect to y. Treat x as constant.

f (x, y)  =  3x² + y² + 5x + 2

∂G/∂y  =  0+2y+0+0

∂G/∂y  =  2y

∂G/∂y at (1, -2)  =  2(-2)

=  -4

Problem 4 :

G(x, y)  =  e^(x+3y) log(x²+y²), (-1, 1)

Solution :

Differentiate with respect to x. Treat y as constant.

∂G/∂x  =  e^(x+3y) 1/(x² + y²) (2x) + log(x² + y²) e^(x+3y) (1)

=  2xe^(x+3y)/(x² + y²) + log(x² + y²) e^(x+3y)

=  e^(x+3y)[2x/(x² + y²)+ log(x² + y²)]

∂G/∂x at (-1, 1)  =  e^(-1+3)[2(-1)/((-1)²+1²)+ log((-1)²+1²)]

=  e²[(-2/2) + log2]

=  e²[log 2 - 1]

Differentiate with respect to y. Treat x as constant.

∂G/∂y  =  e^(x+3y) (1/(x² + y²) (2y) + log(x² + y²) e^(x+3y) (3)

=  2ye^(x+3y)/(x² + y²) + 3log(x² + y²) e^(x+3y)

∂G/∂y  =  e^(x+3y)[2y/(x² + y²) + 3 log(x² + y²)]

∂G/∂y at (-1, 1)  =  e^(-1+3(1))[2(1)/((-1)²+1²)+3log((-1)²+1²)]

=  e²[1 + 3log2]

=  e²[1 + log2³]

=  e²[1 + log8]

Problem 5 :

Find ∂F/∂x, ∂F/∂y for the following functions :

a)  f(x, y) = x cos x sin y

b)  f(x, y) = e^{xy^2}

c)  f(x, y) = (x² + y²) ln (x² + y²)

Solution :

a)  f(x, y) = x cos x sin y

∂F/∂x = (1) cos x sin y + x (-sin x) sin y

= (1) cos x sin y - x sin x sin y

∂F/∂y = x cos x cos y

b)  f(x, y) = e^{xy^2}

∂F/∂x = y² e^{xy^2}

∂F/∂y = 2xy e^{xy^2}

c)  f(x, y) = (x² + y²) ln (x² + y²)

∂F/∂x = 2x ln (x² + y²) + 2x

∂F/∂y = 2y ln (x² + y²) + 2y

Problem 6 :

If u = log (x² + y² + z²) show that 

x(∂²u/∂y ∂z) =  y(∂²u/∂z ∂x) = z(∂²u/∂y ∂x)

Solution :

u = log (x² + y² + z²)

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

(∂u/∂x) = [1/(x² + y² + z²)] 2x

= [2x/(x² + y² + z²)] 

Differentiating with respect to y and considering x and z as constant.

(∂²u/∂y ∂x) = [-4xy/(x² + y² + z²)²]

Multiplying both sides by z,

z(∂²u/∂y ∂x) = [-4xyz/(x² + y² + z²)²]

u = log (x² + y² + z²)

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

(∂u/∂x) = [1/(x² + y² + z²)] 2x

= [2x/(x² + y² + z²)] 

Differentiating with respect to z and considering y and z as constant.

(∂²u/∂z ∂x) = [-4xz/(x² + y² + z²)²]

Multiplying both sides by y,

y(∂²u/∂z ∂x) = [-4xyz/(x² + y² + z²)²]

u = log (x² + y² + z²)

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

(∂u/∂z) = [1/(x² + y² + z²)] 2z

= [2z/(x² + y² + z²)] 

Differentiating with respect to y and considering z and z as constant.

(∂²u/∂y ∂z) = [-4zy/(x² + y² + z²)²]

Multiplying both sides by x,

x(∂²u/∂z ∂x) = [-4xyz/(x² + y² + z²)²]

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