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Question
For the following functions find the fx, and fy and show that fxy = fyx
f(x, y) = `(3x)/(y + sinx)`
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Solution
fx = `((y + sinx)[3] - 3x[0 + cos x])/(y + sin x)^2`
= `(3y + 3sinx - 3xcosx)/(y + sinx)^2`
fy = `((y + sin x)[0] - 3x[1 + 0])/(y + sinx)^2`
= `(- 3x)/(y + sinx)^2`
`(del^2"f")/(delxdely) = del/(delx)[(- 3x)/(y + sinx)^2]`
= `((y + sinx)^2 [- 3] - (- 3x)2(y + sinx)[0 + cosx])/(y + sinx)^4`
= `(- 3(y + sinx)^2 + 6x cosx(y + sinx))/(y + sinx)^4` ........(1)
`(del^2"f")/(delydelx) = del/(dely) [(3y + 3sinx - 3x cosx)/(y + sinx)^2]`
= `((y + sinx)^2[3] - (3y + 3sinx)2(y + sinx)(0 + cosx))/(y + sinx)^4`
= `(-3(y + sinx)^2 + 6x cos x(y + sinx))/(y + sin x)^4` ........(2)
From (1) and (2)
⇒ `(del^2"f")/(delxdely) = (del^2"f")/(delydelx)`
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