Question

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

Answer 1

thanks for questioning.

Answer 2

Answer:

$\frac{d}{dx}\sqrt{cos x} +\sqrt{cos\sqrt{x} } = \frac{-sin x}{2\sqrt{cos x} }-\frac{sin\sqrt{x}}{4\sqrt{x} \sqrt{cos \sqrt{x} } }$

Step-by-step explanation:

As we know that while differentiation we first do the differentiation of power,then the function

let

$y=\sqrt{cos x} +\sqrt{cos\sqrt{x} } \\ \\ y= y1+y2\\ \\ y1=\sqrt{cos x}\\ \\ y2=\sqrt{cos \sqrt{x} }\\ \\ \frac{dy1}{dx}=\frac{1}{2\sqrt{cos x} } .\frac{d cos x}{dx} \\ \\ =\frac{-sin x}{2\sqrt{cos x} }\\ \\ \\\frac{dy2}{dx} =\frac{1}{2\sqrt{cos \sqrt{x} } }.\frac{d cos\sqrt{x} }{dx}\\ \\ =\frac{1}{2\sqrt{cos \sqrt{x} } }.(-sin\sqrt{x} ) \frac{d\sqrt{x} }{dx}\\ \\ =\frac{1}{2\sqrt{cos \sqrt{x} } }.(-sin\sqrt{x} ) \frac{1}{2\sqrt{x} }$

So

$\frac{dy}{dx}=\frac{-sin x}{2\sqrt{cos x} }+\frac{-sin\sqrt{x}}{4\sqrt{x} \sqrt{cos \sqrt{x} } } \\ \\\\=\frac{-sin x}{2\sqrt{cos x} }-\frac{sin\sqrt{x}}{4\sqrt{x} \sqrt{cos \sqrt{x} } }$