Question

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

This question was posed to me by my college director while I was bunking the class.

Origin of the question is First Order PDE in division Partial Differential Equations of Engineering Mathematics

Answer 1

The correct answer is (c) \(\frac{-1}{2\sqrt x}×sin⁡(\sqrt x+\sqrt y)\)

Best explanation: Here we use partial differentiation.

\(u=cos⁡(\sqrt x+\sqrt y)\)

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

\(\frac{\partial z}{\partial x}=\frac{-1}{2\sqrt x}×sin⁡(\sqrt x+\sqrt y)\).