Question
Download Solution PDFযদি u = xyz, v = xy + yz + zx, w = x + y + z হয়, তাহলে \(\frac{\partial(u, v, w)}{\partial(x, y, z)}\) হল:
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFধারণা:
জ্যাকোবিয়ান \(\frac{\partial(u, v, w)}{\partial(x, y, z)}\) দ্বারা দেওয়া হয়:
\(J=\begin{vmatrix} {\partial u\over \partial x} & {\partial u\over \partial y} & {\partial u\over \partial z}\\ {\partial v\over \partial x} & {\partial v\over \partial y} & {\partial v\over \partial z}\\ {\partial w\over \partial x} &{\partial w\over \partial y} & {\partial w\over \partial z} \end{vmatrix}\)
গণনা:
দেওয়া আছে, u = xyz, v = xy + yz + zx, w = x + y + z
\(J=\begin{vmatrix} {\partial (xyz)\over \partial x} & {\partial (xyz)\over \partial y} & {\partial (xyz)\over \partial z}\\ {\partial (xy + yz + zx)\over \partial x} & {\partial (xy + yz + zx)\over \partial y} & {\partial (xy + yz + zx)\over \partial z}\\ {\partial (x + y + z)\over \partial x} &{\partial (x + y + z)\over \partial y} & {\partial (x + y + z)\over \partial z} \end{vmatrix}\)
\(J=\begin{vmatrix} yz & xz & xy\\ y+z & x+z & x+y\\ 1 & 1 & 1 \end{vmatrix}\)
J = yz(x+z-x-y) - xz(y+z-x-y) + xy(y+z-x-z)
J = yz(z-y) - xz(z-x) + xy(y-x)
J = \(yz^2-y^2z-xz^2+x^2z+xy^2-x^2y\)
J = (x - y)(y - z)(z - x)
Last updated on Oct 21, 2023
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