calculate the derivative of the following function.\ny = \\sqrt{2 + \\cot^{2}x}\n\\frac{dy}{dx}=\\square

Explanation:

Step1: Rewrite the function

Let $u = 2+\cot^{2}x$, then $y = \sqrt{u}=u^{\frac{1}{2}}$.

Step2: Find $\frac{dy}{du}$

Using the power - rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

Step3: Find $\frac{du}{dx}$

We know that $\frac{d}{dx}(2) = 0$ and $\frac{d}{dx}(\cot^{2}x)$. Let $v=\cot x$, then $\cot^{2}x = v^{2}$. First, $\frac{dv}{dx}=-\csc^{2}x$. Then, by the chain - rule $\frac{d}{dx}(v^{2}) = 2v\frac{dv}{dx}=2\cot x(-\csc^{2}x)=- 2\cot x\csc^{2}x$. So, $\frac{du}{dx}=-2\cot x\csc^{2}x$.

Step4: Use the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$ and $\frac{du}{dx}=-2\cot x\csc^{2}x$ into the chain - rule. Since $u = 2+\cot^{2}x$, we get $\frac{dy}{dx}=\frac{1}{2\sqrt{2 + \cot^{2}x}}\cdot(-2\cot x\csc^{2}x)=-\frac{\cot x\csc^{2}x}{\sqrt{2+\cot^{2}x}}$.

Answer:

$-\frac{\cot x\csc^{2}x}{\sqrt{2+\cot^{2}x}}$