Question

find $\frac{dy}{dx}$ for $y = \frac{\csc x}{1 + \csc x}$. $\frac{dy}{dx} = $

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 \((-\csc x\cot x)(1 + \csc x)-\csc x(-\csc x\cot x)\):

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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:

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