If z-cosx-y-sinx-y- then x- u- x-y- u- y-

The correct option is C 0 ∂ u∂ #10;x= sin ( xy ) ( 1y )+cos ( #10;xy ) 1y ∂ #10;u∂ y= sin ( xy ) ( xy^2 #10; )+cos ( xy ) ( xy^2 ) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Equations of the Form acosθ+bsinθ = c

If z=cosx/y...

Question

If $z = cos (\frac{x}{y}) + sin (\frac{x}{y})$ ,

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

z

No worries! We‘ve got your back. Try BYJU‘S free classes today!

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 z

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

u = \frac{y}{z} + \frac{z}{x}

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = 0

u = (x^{3} + y^{3} + z^{3}) (x + y + z)

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} = ?

u = \frac{1}{\sqrt{x^{2} + y^{2} + z^{2}}}

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} =

u = ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & 1 x & y & z x^{2} & y^{2} & z^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z} =

u = sin ⎧ ⎪ ⎨ ⎪ ⎩ \frac{π (x^{2} + y^{2} + z x)^{\frac{1}{2}}}{2 (x^{2} + x y + 2 y z + z^{2})^{\frac{1}{3}}} ⎫ ⎪ ⎬ ⎪ ⎭

x \frac{\partial u}{\partial x} + y \frac{\partial u}{\partial y} + z \frac{\partial u}{\partial z}

x = 0; y = 1; z = 2

Join BYJU'S Learning Program