91Ó°ÊÓ

We have \(u(x) = \cot{x}\) and \(v(x) = 1 + \csc{x}\). We need to find the derivative of these functions. Recall that \(\cot{x} = \frac{\cos{x}}{\sin{x}}\). Using the quotient rule, we get: $$u'(x) = \frac{d(\cot x)}{dx} = \frac{-\sin{x}\cos{x} - \cos{x}\sin{x}}{\sin^2{x}} = \frac{-2\sin{x}\cos{x}}{\sin^2{x}}$$ Also, \(\csc{x} = \frac{1}{\sin{x}}\), so to find the derivative of \(v(x)\), we first find the derivative of \(\csc{x}\): $$\frac{d(\csc x)}{dx} = \frac{-\cos{x}}{\sin^2{x}}$$ Now, we find the derivative of \(v(x) = 1 + \csc{x}\), which is: $$v'(x) = \frac{d(1 + \csc x)}{dx} = 0 + \frac{-\cos{x}}{\sin^2{x}} = \frac{-\cos{x}}{\sin^2{x}}$$

Now that we have the derivatives of \(u(x)\) and \(v(x)\), we can apply the quotient rule to find the derivative of the given function: $$\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$$ Plugging in the expressions for \(u'(x)\), \(v(x)\), and \(v'(x)\), we get: $$\frac{d}{dx}\left(\frac{\cot x}{1+\csc x}\right) = \frac{\frac{-2\sin{x}\cos{x}}{\sin^2{x}} \cdot (1 + \csc{x}) - \cot{x} \cdot \frac{-\cos{x}}{\sin^2{x}}}{(1+\csc x)^2}$$

Our goal now is to simplify the expression obtained in step 2. Let's work on the numerator first: $$N = (-2\sin{x}\cos{x})(1 + \frac{1}{\sin{x}}) + \frac{\cos^2{x}}{\sin{x}}\cot{x}$$ $$N = -2\sin{x}\cos{x} -\frac{2\cos{x}}{\sin{x}} + \frac{\cos^2{x}\cos{x}}{\sin^2{x}}$$ Now the denominator, we have: $$D = (1+\csc{x})^2 = (1 + \frac{1}{\sin{x}})^2 = \frac{(\sin{x} + 1)^2}{\sin^2{x}}$$ Hence, the derivative is: $$\frac{d}{dx}\left(\frac{\cot x}{1+\csc x}\right) = \frac{N}{D} = \frac{-2\sin{x}\cos{x} -\frac{2\cos{x}}{\sin{x}} + \frac{\cos^2{x}\cos{x}}{\sin^2{x}}}{\frac{(\sin{x} + 1)^2}{\sin^2{x}}}$$ Finally, we get: $$\frac{d}{dx}\left(\frac{\cot x}{1+\csc x}\right) = \frac{-2\sin{x}\cos{x}\sin^2{x} - 2\cos{x}\sin{x} + \cos^2{x}\cos{x}}{(\sin{x} + 1)^2}$$