If u-x3-3x2y-y3then x-2u- x2-y-2u- x- y-

The correct option is A 6(x^2 2xy) #10; ∂ u∂ x=3x^2 6xy #10; #10; ∂^2u∂ x^2=6x 6y #10; #10; ∂ ^2u∂ y∂ x= 6x #10; #10; x∂^ ...

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(a-b)^2 Expansion and Visualisation

If u=x3-3x2...

Question

If $u = x^{3} - 3 x^{2} y + y^{3}$

then $x \frac{\partial^{2} u}{\partial x^{2}} + y \frac{\partial^{2} u}{\partial x \partial y} = ?$

6 (x^{2} - 2 x y)

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6 x (x - y)

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6 (x - 2 y)

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6 (2 y - x)

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Solution

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Similar questions

u = x^{3} - 3 x^{2} y + y^{3}

x \frac{\partial^{2} u}{\partial x \partial y} + y \frac{\partial^{2} u}{\partial y^{2}} =

u = {tan}^{- 1} (\frac{x^{3} + y^{3}}{x - y})

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?

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u = \frac{e^{x / y} sin (y / x)}{x^{3}}

x^{2} \frac{\partial^{2} u}{\partial x^{2}} + 2 x y \frac{\partial^{2} u}{\partial x \partial y} + y^{2} \frac{\partial^{2} u}{\partial y^{2}} = ?

u = \frac{x}{y^{2}} - \frac{y}{x^{2}}

\frac{\partial^{2} u}{\partial x \partial y} = \frac{\partial^{2} u}{\partial y \partial x}

I

u = \frac{x^{2} y^{2}}{x + y}

⎛ ⎜ ⎜ ⎜ ⎜ ⎝ \frac{x \frac{\partial^{2} u}{\partial x^{2}} + y \frac{\partial^{2} u}{\partial x \partial y}}{\frac{\partial u}{\partial x}} ⎞ ⎟ ⎟ ⎟ ⎟ ⎠

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