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If \( u=\log \left(x^{2}+y^{2}+z^{2}\right) \), then the value of \( x u_{x}+y u_{y}+z u_{z} \) is equal to |
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Answer» u = log(x2 + y2 + z2) ⇒ eu = x2 + y2 + z2 For \(u_x = \frac{\partial u}{\partial x}, e^u \frac{\partial u}{\partial x}= 2x\) \(\frac{\partial u}{\partial x} = \frac{2x}{e^u} = 2xe^{-u}\) Similarly \(\frac{\partial u}{\partial y} = 2ye^{-u}\) & \(\frac{\partial u}{\partial z} = 2ze^{-u}\) Now, xux + yuy + zuz = 2x2e-u + 2y2e-u + 2z2e-4 = 2e-u (x2 + y2 + z2) = 2e-u eu = 2 |
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