This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cotangent function. The function is R(x) = cot(√3 x), so let u = √3 x with u' = √3. The derivative of cot(u) is -csc²(u) multiplied by u', giving -√3 csc²(√3 x). This incorporates the irrational constant. A tempting distractor is choice C, which omits the √3 factor. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem requires differentiating cotangent with a linear argument involving π, testing reciprocal trig derivatives and chain rule precision. The derivative of cot(u) is -csc²(u)·u', and for w(x) = cot(πx), we have u = πx, giving u' = π. Thus w'(x) = -csc²(πx)·π = -π·csc²(πx). A common error would be confusing cotangent's derivative with tangent's and writing π·sec²(πx), but cotangent derivatives always involve cosecant squared, not secant squared. Remember to treat π as a constant coefficient when applying the chain rule, just like any other numerical coefficient.

This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. Y(x) = 3 csc(x), derivative is 3 * (-csc(x) cot(x)) = -3 csc(x) cot(x). The constant 3 multiplies the standard derivative. No inner function beyond x. Choice D uses csc², which is for cot, not csc, a mix-up distractor. Use mnemonic devices like 'csc cot' for csc and 'csc squared' for cot to avoid confusion.

This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. For E(x) = csc(x² - x + 4), the derivative is -csc(u) cot(u) u' where u = x² - x + 4. Then u' = 2x - 1. So E'(x) = -csc(x² - x + 4) cot(x² - x + 4) * (2x - 1). Choice E substitutes csc² incorrectly, tempting if mixing with cot's derivative, but csc uses csc cot with a negative. Consistently recall that csc and cot derivatives both have negatives, but their forms differ.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cosecant function. The function is b(x) = csc(1 - 3x), so let u = 1 - 3x with u' = -3. The derivative of csc(u) is -csc(u) cot(u) multiplied by u', resulting in 3 csc(1 - 3x) cot(1 - 3x) due to double negatives. This shows sign handling in chain rule. A tempting distractor is choice A, which keeps a negative by mishandling signs. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. For B(x) = $cot(e^x$), use -csc²(u) u' with u = $e^x$. u' = $e^x$. So B'(x) = $-csc²(e^x$) * $e^x$ = $-e^x$ $csc²(e^x$). Choice E confuses with csc cot form, which is for csc, not cot, but cot uses csc². Distinguish between the derivative patterns: csc² for cot, sec tan for sec, and csc cot for csc.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cotangent function. The function is p(x) = cot(x² + 1), so let u = x² + 1 with u' = 2x. The derivative of cot(u) is -csc²(u) multiplied by u', yielding -2x csc²(x² + 1). This emphasizes the chain rule for quadratic inners. A tempting distractor is choice E, which mixes cotangent with cosecant derivatives incorrectly. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions, specifically the cosecant function. The function is T(x) = csc(x²), so let u = x² with u' = 2x. The derivative of csc(u) is -csc(u) cot(u) multiplied by u', yielding -2x csc(x²) cot(x²). This demonstrates the negative sign inherent in the cosecant derivative combined with the chain rule. A tempting distractor is choice C, which forgets the 2x factor from the chain rule, ignoring the inner quadratic. When differentiating composite trigonometric functions, always multiply by the derivative of the inner function to apply the chain rule correctly.

This problem tests the skill of differentiating reciprocal trigonometric functions like cotangent, secant, and cosecant. For Z(x) = cot(ln x), derivative is -csc²(u) u' with u = ln x. u' = 1/x. So Z'(x) = -csc²(ln x) * (1/x) = - (1/x) csc²(ln x). Choice C omits 1/x, a chain rule error, but ln x requires it. Always identify if the argument is composite and include its derivative.

This problem involves differentiating the cosecant function with a linear argument. The derivative of csc(u) is -csc(u)cot(u) times the derivative of u. For f(x) = csc(2x-1), we apply the chain rule: f'(x) = -csc(2x-1)cot(2x-1) · 2 = -2csc(2x-1)cot(2x-1). A tempting mistake is to forget the negative sign (choice A), which comes from the fundamental derivative formula for cosecant. The key strategy is to remember that csc and cot derivatives always carry a negative sign, unlike sec and tan derivatives.