Find $\frac{\partial u}{\partial x}$ where $u=cos⁡(\sqrt x+\sqrt y)$.

Question

Find $\frac{\partial u}{\partial x}$  where $u=cos⁡(\sqrt x+\sqrt y)$.

TuteeHUB · Accepted Answer

$\frac{-1}{\sqrt x}×sin⁡(\sqrt x+\sqrt y)$

TuteeHUB · Answer

$\frac{-1}{2\sqrt x}×tan⁡(\sqrt x+\sqrt y)$

TuteeHUB · Answer

$\frac{-1}{2\sqrt x}×cos⁡(\sqrt x+\sqrt y)$

TuteeHUB · Answer

$\frac{-1}{2\sqrt x}×sin⁡(\sqrt x+\sqrt y)$

TuteeHUB · Answer

$\frac{-1}{\sqrt x}×sin⁡(\sqrt x+\sqrt y)$

1.	Find \(\frac{\partial u}{\partial x}\) where \(u=cos⁡(\sqrt x+\sqrt y)\).
A.	\(\frac{-1}{2\sqrt x}×tan⁡(\sqrt x+\sqrt y)\)
B.	\(\frac{-1}{2\sqrt x}×cos⁡(\sqrt x+\sqrt y)\)
C.	\(\frac{-1}{2\sqrt x}×sin⁡(\sqrt x+\sqrt y)\)
D.	\(\frac{-1}{\sqrt x}×sin⁡(\sqrt x+\sqrt y)\)
Answer» D. \(\frac{-1}{\sqrt x}×sin⁡(\sqrt x+\sqrt y)\)

Find \(\frac{\partial u}{\partial x}\) where \(u=cos⁡(\sqrt x+\sqrt y)\).