# PARTIAL DERIVATIVE OF A FUNCTION OF TWO VARIABLES

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 .

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)+3log(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]

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