Question

$u {}^{3} = xyz \: \: \frac{1}{v} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z \: \: } \: \: w {}^{2} = x {}^{2} + y {}^{2} + z {}^{2} \: \: then \: prove \: that \: \: \\ \frac{d(u \: v \: w)}{d(x \: y \: z) \: } = \frac{v(y - z) \: (z - x) \: (x - y) \: (x + y + z)}{3u {}^{2}w(yz + zx + xy) }$
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Answer 1

Answer:

what is that tex]u {}^{3} = xyz \: \: \frac{1}{v} = \frac{1}{x} + \frac{1}{y} + \frac{1}{z \: \: } \: \: w {}^{2} = x {}^{2} + y {}^{2} + z {}^{2} \: \: then \: prove \: that \: \: \\ \frac{d(u \: v \: w)}{d(x \: y \: z) \: } = \frac{v(y - z) \: (z - x) \: (x - y) \: (x + y + z)}{3u {}^{2}w(yz + zx + xy) } [/tex]