Question

Differentiate the function w.r.t.x. : $\sqrt{\sin x + \sqrt{\cos x}}$

Answer 1

nice question another

Answer 2

Answer:

$\frac{d}{dx}\sqrt{sin\:x+\sqrt{cos\:x} }= \frac{1}{2\sqrt{sin\:x+\sqrt{cos\:x} }}.(cos\:x-\frac{sin\:x}{2\sqrt{cos\:x} })$

Step-by-step explanation:

As we know that, first differentiate the power,than by chain rule differentiate remaining function

$y=\sqrt{sin\:x+\sqrt{cos\:x} }\\ \\ \frac{dy}{dx} =\frac{1}{2\sqrt{sin\:x+\sqrt{cos\:x} }}.\frac{d}{dx}(sin\:x+\sqrt{cos\:x} )\\ \\ \\= \frac{1}{2\sqrt{sin\:x+\sqrt{cos\:x} }}.(cos\:x+\frac{1}{2\sqrt{cos\:x} }.\frac{d\:cos\:x}{dx}) \\ \\ \\ =\frac{1}{2\sqrt{sin\:x+\sqrt{cos\:x} }}.(cos\:x+\frac{-sin\:x}{2\sqrt{cos\:x} })\\ \\ \\ = \frac{1}{2\sqrt{sin\:x+\sqrt{cos\:x} }}.(cos\:x-\frac{sin\:x}{2\sqrt{cos\:x} })$