calculate the derivative of the following function. y = \\sqrt{6 + \\cot^{2}x} \\frac{dy}{dx}=\\square

Explanation:

Step1: Let $u = 6+\cot^{2}x$

$y=\sqrt{u}=u^{\frac{1}{2}}$

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 $\frac{du}{dx}$

Let $v = \cot x$, then $u = 6 + v^{2}$. First, $\frac{d}{dv}(6 + v^{2})=2v$ and $\frac{dv}{dx}=-\csc^{2}x$. By the chain - rule $\frac{du}{dx}=\frac{d}{dv}(6 + v^{2})\cdot\frac{dv}{dx}=2v\cdot(-\csc^{2}x)=-2\cot x\csc^{2}x$

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

Substitute $u = 6+\cot^{2}x$ and $\frac{du}{dx}=-2\cot x\csc^{2}x$ and $\frac{dy}{du}=\frac{1}{2\sqrt{u}}$ into the chain - rule formula: $\frac{dy}{dx}=\frac{1}{2\sqrt{6+\cot^{2}x}}\cdot(-2\cot x\csc^{2}x)$

Answer:

$-\frac{\cot x\csc^{2}x}{\sqrt{6 + \cot^{2}x}}$