Verify the following derivative formula using the Quotient Rule. \[ \frac{d}{d x}(\cot x)=-\csc ^{2} x \] Choose the correct sequence of steps below that verifies the given derivative formula. A. \[ \begin{aligned} \frac{d}{d x}(\cot x) & =\frac{d}{d x}\left(\frac{\cos x}{\sin x}\right) \\ & =\frac{\sin x(-\sin x)-\cos x \cos x}{\sin ^{2} x} \\ & =-\csc ^{2} x \end{aligned} \] \( B \). \[ \begin{aligned} \frac{d}{d x}(\cot x) & =\frac{d}{d x}\left(\frac{\cos x}{\sin x}\right) \\ & =\frac{\cos x(-\cos x)-\sin x \sin x}{\cos ^{2} x} \\ & =-\csc ^{2} x \end{aligned} \]

To verify the derivative of \(\cot x\) using the Quotient Rule, let's analyze option A: Starting with \(\cot x = \frac{\cos x}{\sin x}\), we apply the Quotient Rule: \[ \frac{d}{d x}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \] where \(u = \cos x\) and \(v = \sin x\). Calculating: - \(u' = -\sin x\) and \(v' = \cos x\). - Therefore, the derivative is: \[ \frac{d}{d x}(\cot x) = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x. \] So, option A is indeed the correct sequence that verifies the formula! Now, let's look at option B: This option tries to apply the Quotient Rule, but it mistakenly replaces the expressions leading to incorrect terms in the derivative calculation. Incorrectly stating \(u' = \cos x\) instead of \(-\sin x\) means it ultimately fails to derive the correct result, thus validating that A is the only correct sequence.