Question

Let u = x cos y + y cos x. Verify `(del^2"u")/(delxdely) = (del^"u")/(del"y"del"x")`

Answer 1

u = x cos y + y cos x

Differentiating partially with respect to y, we get,

`(delu)/(dely) = del/(dely) (x cos y) + del/(dely) (y cos x)`

`= x del/(del y) (cos y) + cos x ddel/(del y) (y)`

= x(-sin y) + cos x

Again differentiating partially with respect to x, we get

`del/(delx) ((delu)/(dely)) = del/(delx) (- x sin y) + del/(delx) (cos x)`

`= del/(delx) (- x sin y) + del/(delx) (cos x)`

`= - sin y del/(delx) (x) + (- sin x)`

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

= -sin y – sin x ……… (1)

Now u = x cos y + y cos x

Differentiating partially with respect to x we get

`(delu)/(delx) = cos y del/(delx) (x) + y del/(delx) (cos x)`

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

= cos y – y sin x

Again differentiating partially with respect to y we get,

`del/(dely) ((delu)/(dely)) = del/(dely)  (cos y - y sin x)`

`= del/(dely) (cos y) - del/(dely) (y sin x)`

= -sin y – sin x `del/(dely)`(y)

= -sin y – sin x (1)

= -sin y – sin x ………(2)

From (1) and (2),

`(del^2"u")/(delxdely) = (del^"u")/(del"y"del"x")`

Hence verified.