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Question Number 41028 by Choudharyvishal155@gmail.com last updated on 31/Jul/18

if 1/u = (√((x2 + y2 +z2))) then x∂u/∂x + y∂u/∂y + z∂u/∂z = ?

if1/u=(x2+y2+z2)thenx∂u/∂x+y∂u/∂y+z∂u/∂z=?

Answered by tanmay.chaudhury50@gmail.com last updated on 31/Jul/18

u=(1/(√(x^2 +y^2 +z^2 ))) so (∂u/∂x)=((−1)/2)(x^2 +y^2 +z^2 )^((−3)/2) ×2x x(∂u/∂x)+y(∂u/∂y)+z(∂u/∂z)=−(x^2 +y^2 +z^2 )(x^2 +y^2 +z^2 )^((−3)/2) =−(x^2 +y^2 +z^2 )^((−1)/2) =−(1/(√(x^2 +y^2 +z^2 ))) =−u

u=1x2+y2+z2so∂u∂x=−12(x2+y2+z2)−32×2xx∂u∂x+y∂u∂y+z∂u∂z=−(x2+y2+z2)(x2+y2+z2)−32=−(x2+y2+z2)−12=−1x2+y2+z2=−u

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