find $\\frac{dy}{dx}$ for $y = \\frac{\\csc x}{1 + \\csc x}$. $\\frac{dy}{dx} = $
find $\\frac{dy}{dx}$ for $y = \\frac{\\csc x}{1 + \\csc x}$. $\\frac{dy}{dx} = $
Answer
Explanation:
Step1: Apply the 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 x), (u^\prime=-\csc x\cot x), (v = 1+\csc x), and (v^\prime=-\csc x\cot x). [ \frac{dy}{dx}=\frac{(-\csc x\cot x)(1 + \csc x)-\csc x(-\csc x\cot x)}{(1 + \csc x)^{2}} ]
Step2: Expand the numerator
Expand ((-\csc x\cot x)(1 + \csc x)-\csc x(-\csc x\cot x)): [ \begin{align*} &(-\csc x\cot x-\csc^{2}x\cot x)+\csc^{2}x\cot x\ =&-\csc x\cot x \end{align*} ]
Step3: Simplify the derivative
Since the numerator is (-\csc x\cot x) and the denominator is ((1 + \csc x)^{2}), the derivative (\frac{dy}{dx}=\frac{-\csc x\cot x}{(1 + \csc x)^{2}})
Answer:
(\frac{-\csc x\cot x}{(1 + \csc x)^{2}})