find the derivative of the following function.\ny = \\frac{\\csc w}{1 + \\csc w}\n\\frac{dy}{dw}=\\square

Explanation:

Step1: Recall quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = \csc w$, $v=1 + \csc w$.

Step2: Find $u^\prime$ and $v^\prime$

The derivative of $\csc w$ is $-\csc w\cot w$. So, $u^\prime=-\csc w\cot w$ and $v^\prime=-\csc w\cot w$.

Step3: Apply the quotient - rule

[ \begin{align*} \frac{dy}{dw}&=\frac{(-\csc w\cot w)(1 + \csc w)-\csc w(-\csc w\cot w)}{(1 + \csc w)^{2}}\ &=\frac{-\csc w\cot w-\csc^{2}w\cot w+\csc^{2}w\cot w}{(1 + \csc w)^{2}}\ &=\frac{-\csc w\cot w}{(1 + \csc w)^{2}} \end{align*} ]

Answer:

$\frac{-\csc w\cot w}{(1 + \csc w)^{2}}$