If u-xz2-yz2-xy- then - x- u- x-

The correct option is D 0 ∂ #10;u∂ x= 2xz^2+ 1y #10;∂ u∂ y= 2yz^2 xy^2 ∂ #10;u∂ z= 2x^2z^3 2y^2Z^3 ∑ x #10;∂ u∂ x= x∂ u∂ x+y ...

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Byju's Answer

Standard XII

Mathematics

Proof by mathematical induction

If u=xz2+yz...

Question

If $u = {(\frac{x}{z})}^{2} + {(\frac{y}{z})}^{2} + (\frac{x}{y})$ , then $Σ x \frac{\partial u}{\partial x} =$

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Solution

The correct option is D $0$
$\frac{\partial u}{\partial x} = \frac{2 x}{z^{2}} + \frac{1}{y}$
$\frac{\partial u}{\partial y} = \frac{2 y}{z^{2}} - \frac{x}{y^{2}}$
$\frac{\partial u}{\partial z} = \frac{- 2 x^{2}}{z^{3}} - \frac{2 y^{2}}{Z^{3}}$
$\sum x \frac{\partial u}{\partial x} = \frac{x \partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}$
$= 0$

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If u=(xz)2+(yz)2+(xy), then Σx∂u∂x=

The correct option is D 0∂u∂x=2xz2+1y∂u∂y=2yz2−xy2∂u∂z=−2x2z3−2y2Z3∑x∂u∂x=x∂u∂x+y∂u∂y+z∂u∂z =0

If $u = {(\frac{x}{z})}^{2} + {(\frac{y}{z})}^{2} + (\frac{x}{y})$ , then $Σ x \frac{\partial u}{\partial x} =$