Question

Find the derivative of the following functions. $$y=\frac{\cot x}{1+\csc x}$$

Short answer

Answer: The derivative of the given function is \(\frac{dy}{dx} = \frac{-\csc^2 x - \csc^3 x + \cot^2 x \csc x}{(1+\csc x)^2}\).

Step-by-step solution

Determine the derivatives of cotangent and cosecant functions

First, we need to express the function \(y\) in terms of sine and cosine. Recall that \(\cot x = \frac{\cos x}{\sin x}\) and \(\csc x = \frac{1}{\sin x}\). So our function becomes: $$y = \frac{\frac{\cos x}{\sin x}}{1+\frac{1}{\sin x}}$$ The derivatives of sine and cosine functions are as follows: $$\frac{d}{dx} \sin x = \cos x \quad \text{and} \quad \frac{d}{dx} \cos x = -\sin x$$ Using these, we can find the derivatives of cotangent and cosecant functions: $$\frac{d}{dx} \cot x = \frac{d}{dx} \left( \frac{\cos x}{\sin x} \right) = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = -\csc^2 x$$ $$\frac{d}{dx} \csc x = \frac{d}{dx} \left( \frac{1}{\sin x} \right) = -\frac{\cos x}{\sin^2 x} = -\cot x \csc x$$

Apply the quotient rule to find the derivative of the function

Now we will find the derivative of the function, \(y\), using the quotient rule. The quotient rule states: $$\frac{d}{dx} \frac{u}{v} = \frac{vu' - uv'}{v^2}$$ Let \(u\) be \(\cot x\) and \(v\) be \((1 + \csc x)\). $$\frac{dy}{dx} = \frac{(1+ \csc x)(-\csc^2 x) - \cot x (-\cot x \csc x)}{(1+\csc x)^2}$$ Now simplify the derivative.

Simplify the derivative

Simplify the result obtained from the quotient rule: $$\frac{dy}{dx} = \frac{-\csc^2 x - \csc^3 x + \cot^2 x \csc x}{(1+\csc x)^2}$$ Finally, the derivative of the given function is: $$\frac{dy}{dx} = \frac{-\csc^2 x - \csc^3 x + \cot^2 x \csc x}{(1+\csc x)^2}$$