यदि u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2 ह | vedclass

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(A) दिया गया है $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$। $x$ के सापेक्ष अवकलन करने पर: $2u \frac{\partial u}{\partial x} = 2(x - a) \implies u \frac

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$\frac{2}{u}$

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(A) दिया गया है $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$। $x$ के सापेक्ष अवकलन करने पर: $2u \frac{\partial u}{\partial x} = 2(x - a) \implies u \frac{\partial u}{\partial x} = x - a$। पुनः $x$ के सापेक्ष अवकलन करने पर: $u \frac{\partial^2 u}{\partial x^2} + \left(\frac{\partial u}{\partial x}\right)^2 = 1$। $\frac{\partial u}{\partial x} = \frac{x - a}{u}$ रखने पर: $u \frac{\partial^2 u}{\partial x^2} = 1 - \left(\frac{x - a}{u}\right)^2 = 1 - \frac{(x - a)^2}{u^2} = \frac{u^2 - (x - a)^2}{u^2}$। अतः,$\frac{\partial^2 u}{\partial x^2} = \frac{1}{u} - \frac{(x - a)^2}{u^3}$। इसी प्रकार,$\frac{\partial^2 u}{\partial y^2} = \frac{1}{u} - \frac{(y - b)^2}{u^3}$ और $\frac{\partial^2 u}{\partial z^2} = \frac{1}{u} - \frac{(z - c)^2}{u^3}$। इनका योग करने पर: $\sum \frac{\partial^2 u}{\partial x^2} = \frac{3}{u} - \frac{1}{u^3} [(x - a)^2 + (y - b)^2 + (z - c)^2]$। चूंकि $(x - a)^2 + (y - b)^2 + (z - c)^2 = u^2$: $\sum \frac{\partial^2 u}{\partial x^2} = \frac{3}{u} - \frac{u^2}{u^3} = \frac{3}{u} - \frac{1}{u} = \frac{2}{u}$।

यदि $u^2 = (x - a)^2 + (y - b)^2 + (z - c)^2$ है,तो $\sum \frac{\partial^2 u}{\partial x^2} = $

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