FIND fxy and fyx USING PARTIAL DERIVATIVE OF A 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 .

For each of the following functions find the fx and fy and show that fxy  =  fyx

Problem 1 :

f (x, y)  =  3x/(y+sinx)

Solution :

f (x, y)  =  3x/(y+sinx)

Finding fx :

Differentiate with respect to x. Treat y as constant.

u  =  3x and v  =  y+sinx

u'  =  3 and v'  =  0+cosx  ==>  cosx

fx  =  [(y+sinx)(3) - 3x(cosx)]/(y+sinx)2

fx  =  [3y+3sinx - 3xcosx]/(y+sinx)2

Finding fxy :

Differentiate with respect to y. Treat x as constant.

Finding fy :

Differentiate with respect to y. Treat x as constant.

u = 3x and v = y+sinx

u' = 0 and v' = 1+0 ==> 1

fy = [(y+sinx)(0) - 3x(1)]/(y+sinx)2

fy = -3x/(y+sinx)2

Finding fyx :

Differentiate with respect to x. Treat y as constant.

u = -3x, v = (y+sinx)2

u' = -3 and v' = 2((y+sinx))cosx

v' = 2cosx(y+sinx)

= [-3(y+sinx)2 -(-3x) 2((y+sinx))cosx]/(y+sinx)4

= (y+sinx)[-3(y+sinx)+6xcosx]/(y+sinx)4

fyx = [-3y-3sinx+6xcosx]/(y+sinx)3

Problem 2 :

f(x, y)  =  tan-1(x/y)

Solution :

f(x, y)  =  tan-1(x/y)

Finding fx :

Differentiate with respect to x. Treat y as constant.

fx = 1/(1+(x/y)2) (1/y)

f= (1/y)/(y2+x2)/y2

f= y/(x2+y2)

Finding fxy :

Differentiate with respect to y. Treat x as constant.

u = y and v = x2+y2

u' = 1 and v' = 2y

fxy  =  [x2+y2-y(2y)]/(x2+y2)

fxy  =  (x2-y2)/(x2+y2)  ---(1)

f(x, y)  =  tan-1(x/y)

Finding fy :

f(x, y)  =  tan-1(xy-1)

Differentiate with respect to y. Treat x as constant.

fy = [1/1+(x/y)2](-x/y2)

fy = [y2/(x2+y2)](-x/y2)

fy = -x/(x2+y2)

Finding fyx :

Differentiate with respect to x. Treat y as constant.

u = -x and v = x2+y2

u' = -1 and v' = 2x

fyx = [(x2+y2)(-1) - (-x)(2x)]/(x2+y2)2

fyx = [-x2-y2 +2x2]/(x2+y2)2

fyx = (x2-y2)/(x2+y2)2 ---(2)

Problem 3 :

f(x, y) = cos (x2-3xy)

Solution :

Finding fx :

Differentiate with respect to x. Treat y as constant.

fx = sin (x2-3xy)(2x-3y)

f= (2x-3y)sin (x2-3xy)

Finding fxy :

u = 2x-3y and v = sin (x2-3xy)

u' = 0-3 and v' = cos (x2-3xy) (0-3x)

v' = -3xcos (x2-3xy)

fxy = (2x-3y)(-3xcos) (x2-3xy) + sin (x2-3xy)(-3)

fxy = (2x-3y)(-3xcos) (x2-3xy) + sin (x2-3xy)(-3)

fxy = (2x-3y)(-3xcos) (x2-3xy) - 3 sin (x2-3xy)  --(1)

Finding fy :

Differentiate with respect to y. Treat x as constant.

fy = sin (x2-3xy)(0-3x)

f= -3x sin (x2-3xy)

Finding fyx :

Differentiate with respect to x. Treat y as constant.

u = -3x and v = sin (x2-3xy)

u' = -3 and v' = cos (x2-3xy) (2x-3y)

fyx = -3x cos (x2-3xy) (2x-3y) -3 sin (x2-3xy) --(2) 

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